# Differentiable Manifolds: Charts and Atlases That Bring Calculus to Curved Spaces > We introduce charts, transition maps and maximal atlases, build explicit atlases on the sphere, projective space and the torus, and define C^k functions and maps. > https://rikai.mugen-giken.com/en/mathematics/manifolds/smooth-manifolds ## 0. Key points - The definition of a manifold comes in two layers. First, as a topological space it must be **locally homeomorphic to $\mathbb{R}^n$**; second, the maps that glue local coordinates to one another (the transition maps) must be **of class $C^k$**. The first layer carries the shape, the second carries the possibility of differentiating. - There is exactly one reason for demanding that the transition maps be $C^k$: **so that the notion of differentiation defined through coordinates does not depend on the choice of coordinates**. Everything in the definition follows from this single requirement. - The definition adds the Hausdorff condition and second countability. Neither is a consequence of the local data. Dropping them lets pathologies such as the line with two origins appear. - For $S^n$ (stereographic projection), $\mathbb{RP}^n$ (homogeneous coordinates) and $T^n = \mathbb{R}^n/\mathbb{Z}^n$ (integer translations), the atlases can be written out completely. The transition maps are, respectively, the inversion $u \mapsto u/|u|^2$, rational expressions, and translations. - The $C^k$ property of functions and maps on a manifold is defined through coordinate representations, and compatibility makes it independent of the chart chosen. That composites are again $C^k$ also follows from this. - One topological space can carry different differentiable structures (the chart $t \mapsto t^3$ on $\mathbb{R}$). Yet such structures may still be diffeomorphic, so "different structure" and "different manifold" are two distinct assertions. ## 1. Motivation: differentiating on a curved space Differentiation is local in nature. The derivative of a function $f$ at a point $a$ is determined by the behaviour of $f$ on arbitrarily small neighbourhoods of $a$ alone. Yet the differentiation taught in calculus is defined only on open subsets of $\mathbb{R}^n$. We may wish to differentiate a temperature distribution on a sphere, or an electromagnetic field on spacetime; neither the sphere nor spacetime is an open subset of $\mathbb{R}^n$, so the definition does not apply as it stands. Historically, Gauss took the decisive step in his 1827 theory of surfaces. He showed that the quantity measuring how a surface curves (the Gaussian curvature) is determined by lengths measured within the surface, independently of how the surface sits inside $\mathbb{R}^3$ (the *Theorema Egregium*). This is a declaration that geometry can be developed from the internal data of a surface, without looking at it from outside. In 1854, in his inaugural lecture at Göttingen, Riemann extended this standpoint to arbitrary dimensions. The word he used, *Mannigfaltigkeit*, is our "manifold". Once we adopt the intrinsic standpoint, the only tool left to us is **coordinates**. Expressing points of a surface by longitude and latitude turns a function on the surface into a function of two variables, which we can differentiate. But longitude and latitude are only one choice. A different coordinate system yields a different function of two variables. Here the decisive question arises. > Is the property of "being differentiable", defined through coordinates, independent of the choice of coordinates? For it to be independent, the map linking two coordinate systems — the transition map — must itself be differentiable. Indeed, writing $f \circ \varphi^{-1}$ for the expression of $f$ in the coordinates $\varphi$ and $f \circ \psi^{-1}$ for its expression in the coordinates $\psi$, the two are related by $$ f \circ \psi^{-1} = (f \circ \varphi^{-1}) \circ (\varphi \circ \psi^{-1}). $$ If $\varphi \circ \psi^{-1}$ on the right is of class $C^k$, then the chain rule carries "$f \circ \varphi^{-1}$ is $C^k$" over to "$f \circ \psi^{-1}$ is $C^k$". **The definition of a manifold is nothing but a staging of this one line.** The sections below translate this intuition into definitions and verify it on representative spaces. Spaces requiring such preparation are by no means artificial. - **Solution sets of equations**: $S^n = \{x \in \mathbb{R}^{n+1} : |x| = 1\}$ is not an open subset of $\mathbb{R}^{n+1}$, yet near each of its points it can be described by $n$ coordinates. - **Configuration spaces**: the state of a double pendulum is determined by two angles, and the totality of such states is the torus $T^2$. The set of attitudes of a rigid body is identified with $\mathbb{RP}^3$. To write equations of motion on these spaces, we must first fix the meaning of differentiation. - **General relativity**: spacetime has no preferred coordinate system (no inertial frame). Physical laws must therefore be written in a form invariant under changes of coordinates. The Jacobian matrix $\partial x'^{\mu}/\partial x^{\nu}$ appearing in the transformation law of tensors is precisely the derivative of a transition map. The requirement that transition maps be $C^k$ is the minimal premise for speaking of general covariance. In this article we define $C^k$ manifolds as a stage satisfying the requirements above, and go as far as the $C^k$ property of functions and maps on them. Tangent vectors, vector fields and differential forms — the tools with which one actually differentiates on a manifold — are treated in [Tangent spaces and the tangent bundle](/en/mathematics/manifolds/tangent-spaces) and [Vector fields and differential forms](/mathematics/manifolds/differential-forms). ## 2. Preliminaries: what we use from topology and multivariable calculus Let us fix notation and hypotheses. We write $\mathbb{N} = \{1, 2, \ldots\}$, and points of $\mathbb{R}^n$ as $u = (u^1, \ldots, u^n)$ with upper indices (the convention in manifold theory, chosen to agree with the tensor notation of later articles). Here $|u|$ is the Euclidean norm. The language of topological spaces is taken from [Topological spaces: definition and basic notions](/en/mathematics/topology/topological-spaces) and [Continuous maps and homeomorphisms](/en/mathematics/topology/continuous-maps). In particular we use the following. - A topological space $X$ is **Hausdorff** when any two distinct points can be separated by disjoint open sets (). - $X$ is **second countable** when its topology has a countable basis (). - A **homeomorphism** is a bijection continuous in both directions (). - The definitions of the subspace, product and quotient topologies. For a quotient map $\pi \colon X \to X/\!\sim$, a set $W \subset X/\!\sim$ is open $\iff$ $\pi^{-1}(W)$ is open. For details on the separation axioms see [Separation axioms and metrizability](/mathematics/topology/separation-axioms). For differentiation in several variables we assume [Differentiation of functions of several variables and partial derivatives](/mathematics/calculus/multivariable-differentiation) and use the following definition. Let $\Omega \subset \mathbb{R}^n$ be open and $F \colon \Omega \to \mathbb{R}^m$. We say $F$ is of class $C^k$ ($k \in \mathbb{N}$) when every component of $F$ has all partial derivatives up to order $k$ () and all of them are continuous on $\Omega$. Class $C^0$ means continuous, and class $C^\infty$ means of class $C^k$ for every $k$. When $F$ is $C^1$, we write $DF(a) \in \mathbb{R}^{m \times n}$ for its derivative (Jacobian matrix) at $a \in \Omega$. Only two properties are used in an essential way in this article. Let $k \in \mathbb{N} \cup \{\infty\}$. 1. (Locality) Let $\Omega \subset \mathbb{R}^n$ be open and $F \colon \Omega \to \mathbb{R}^m$. If for every $a \in \Omega$ there is an open neighbourhood $W_a \subset \Omega$ of $a$ such that $F|_{W_a}$ is $C^k$, then $F$ is $C^k$ on $\Omega$. 2. (Composition) Let $\Omega \subset \mathbb{R}^n$ and $\Omega' \subset \mathbb{R}^m$ be open, let $F \colon \Omega \to \mathbb{R}^m$ be $C^k$ with $F(\Omega) \subset \Omega'$, and let $G \colon \Omega' \to \mathbb{R}^l$ be $C^k$. Then $G \circ F \colon \Omega \to \mathbb{R}^l$ is $C^k$, and for $k \ge 1$ the chain rule $D(G \circ F)(a) = DG(F(a)) \, DF(a)$ holds at every $a \in \Omega$. Part 1 is immediate from the fact that partial derivatives are determined on a neighbourhood of each point. Part 2 follows from the chain rule () applied inductively: higher partial derivatives can be written using products and composites of the entries of $DF$ and $DG$, hence remain continuous up to order $k$. For detailed proofs see [Differentiation of functions of several variables and partial derivatives](/mathematics/calculus/multivariable-differentiation). ## 3. Charts and atlases ### 3.1. Charts as local coordinate systems Let $M$ be a topological space and $n$ an integer with $n \ge 0$. A pair $(U, \varphi)$ is an **$n$-dimensional chart** (coordinate neighbourhood) of $M$ when 1. $U$ is an open subset of $M$, and 2. $\varphi \colon U \to \varphi(U)$ is a homeomorphism onto some open subset $\varphi(U)$ of $\mathbb{R}^n$. Writing the $i$-th component of $\varphi$ as $x^i \colon U \to \mathbb{R}$, we call $\varphi = (x^1, \ldots, x^n)$ a **local coordinate system** on $U$ and each $x^i$ a **coordinate function**. For $p \in U$, the tuple $\varphi(p) = (x^1(p), \ldots, x^n(p))$ is called the **coordinates** of $p$. When every point of $M$ lies in the domain of some $n$-dimensional chart, $M$ is called an **$n$-dimensional locally Euclidean space**. A chart is a device that identifies part of $M$ with an open subset of $\mathbb{R}^n$. It corresponds to transferring a portion of the globe onto a flat map, whence the name *chart*. Just as no single map covers the whole Earth, a single chart cannot in general cover all of $M$. So we provide several of them and require consistency on the overlaps. ### 3.2. Transition maps and compatibility When the domains of two charts overlap, the overlap carries two sets of coordinates. The map linking them is the transition map. Let $k \in \mathbb{N} \cup \{\infty\}$. Two $n$-dimensional charts $(U, \varphi)$ and $(V, \psi)$ of $M$ are **$C^k$ compatible** when either $U \cap V = \emptyset$, or both $$ \psi \circ \varphi^{-1} \colon \varphi(U \cap V) \longrightarrow \psi(U \cap V), \qquad \varphi \circ \psi^{-1} \colon \psi(U \cap V) \longrightarrow \varphi(U \cap V) $$ are of class $C^k$. These maps are called **transition maps** (change of coordinates). Let us check that this definition makes sense. The set $U \cap V$ is open in $M$, and $\varphi$ is a homeomorphism from $U$ onto $\varphi(U)$, so $\varphi(U \cap V)$ is open in $\varphi(U)$. Since $\varphi(U)$ is itself open in $\mathbb{R}^n$, the set $\varphi(U \cap V)$ is open in $\mathbb{R}^n$; similarly for $\psi(U \cap V)$. Hence $\psi \circ \varphi^{-1}$ is a map from an open subset of $\mathbb{R}^n$ to an open subset of $\mathbb{R}^n$, and asking whether it is $C^k$ makes sense. Note also that $\psi \circ \varphi^{-1}$ and $\varphi \circ \psi^{-1}$ are mutually inverse, so both are homeomorphisms.
M U V U ∩ V φ(U) ⊂ ℝⁿ ψ(V) ⊂ ℝⁿ φ ψ ψ ∘ φ⁻¹ transition map
Let $M$ be a topological space and $k \in \mathbb{N} \cup \{\infty\}$. 1. A family $\mathcal{A} = \{(U_\alpha, \varphi_\alpha)\}_{\alpha \in A}$ of $n$-dimensional charts of $M$ is an **$n$-dimensional $C^k$ atlas** when $\bigcup_{\alpha \in A} U_\alpha = M$ and, for all $\alpha, \beta \in A$, the charts $(U_\alpha, \varphi_\alpha)$ and $(U_\beta, \varphi_\beta)$ are $C^k$ compatible. 2. A $C^k$ atlas $\mathcal{A}$ is **maximal** when every $n$-dimensional chart of $M$ that is $C^k$ compatible with all charts of $\mathcal{A}$ already belongs to $\mathcal{A}$. 3. A maximal $n$-dimensional $C^k$ atlas is called an **$n$-dimensional differentiable structure of class $C^k$** ($C^k$ structure) on $M$. Maximality is required in order to identify different atlases that determine the same differentiable structure. For instance, as we shall see, the sphere $S^n$ carries an atlas of two charts given by stereographic projection and an atlas of $2n+2$ charts given by graph representations over hemispheres, and these give the same notion of differentiation. Passing to maximal atlases, the two become literally the same set. ### 3.3. Existence and uniqueness of the maximal atlas Writing down a maximal atlas directly is impossible in practice. The next proposition guarantees that it suffices to give one small atlas. The heart of the proof is the following lemma. Let $\mathcal{A}$ be an $n$-dimensional $C^k$ atlas of a topological space $M$. If two $n$-dimensional charts $(U, \varphi)$ and $(V, \psi)$ of $M$ are both $C^k$ compatible with every chart of $\mathcal{A}$, then $(U, \varphi)$ and $(V, \psi)$ are $C^k$ compatible with each other. If $U \cap V = \emptyset$ the claim holds by definition, so assume $U \cap V \ne \emptyset$. It suffices to show that $\psi \circ \varphi^{-1}$ is $C^k$ on $\varphi(U \cap V)$; the other direction follows by exchanging the roles of $\varphi$ and $\psi$. By part 1 (locality) of , it is enough to prove that the map is $C^k$ on a neighbourhood of each point of $\varphi(U \cap V)$. So take an arbitrary $a \in \varphi(U \cap V)$ and put $p = \varphi^{-1}(a) \in U \cap V$. Since $\mathcal{A}$ covers $M$, there is a chart $(W, \chi) \in \mathcal{A}$ with $p \in W$. The set $U \cap V \cap W$ is open in $M$ and contains $p$, and $\varphi$ is a homeomorphism, so $\varphi(U \cap V \cap W)$ is an open subset of $\mathbb{R}^n$ containing $a$. On this open set we have $$ \psi \circ \varphi^{-1} = (\psi \circ \chi^{-1}) \circ (\chi \circ \varphi^{-1}). $$ Indeed, for $s \in \varphi(U \cap V \cap W)$ we have $q = \varphi^{-1}(s) \in U \cap V \cap W$ and $\chi \circ \varphi^{-1}(s) = \chi(q) \in \chi(U \cap V \cap W) \subset \chi(V \cap W)$, so the right-hand side is defined, and its value is $\psi(q) = \psi \circ \varphi^{-1}(s)$. Now $\chi \circ \varphi^{-1}$ is $C^k$ on $\varphi(U \cap W)$, because by hypothesis $(U, \varphi)$ is $C^k$ compatible with $(W, \chi) \in \mathcal{A}$. Likewise $\psi \circ \chi^{-1}$ is $C^k$ on $\chi(V \cap W)$, because $(V, \psi)$ is $C^k$ compatible with $(W, \chi)$. Hence by part 2 (composition) of , the map $\psi \circ \varphi^{-1}$ is $C^k$ on $\varphi(U \cap V \cap W)$. Since $a$ was arbitrary, locality gives that $\psi \circ \varphi^{-1}$ is $C^k$ on all of $\varphi(U \cap V)$. Let $M$ be a topological space and $\mathcal{A}$ an $n$-dimensional $C^k$ atlas of $M$. Then there is exactly one maximal $n$-dimensional $C^k$ atlas containing $\mathcal{A}$, namely $$ \mathcal{A}^{*} = \{ (U, \varphi) : (U,\varphi) \text{ is an } n \text{-dimensional chart of } M \text{ that is } C^k \text{ compatible with every chart of } \mathcal{A} \}. $$ **(1) $\mathcal{A} \subset \mathcal{A}^{*}$.** Since $\mathcal{A}$ is an atlas, any two of its charts are $C^k$ compatible. Hence every chart of $\mathcal{A}$ satisfies the defining condition of $\mathcal{A}^{*}$. In particular the domains of the charts of $\mathcal{A}^{*}$ cover $M$. **(2) $\mathcal{A}^{*}$ is an atlas.** Covering was shown in (1). If $(U, \varphi), (V, \psi) \in \mathcal{A}^{*}$, then both are $C^k$ compatible with every chart of $\mathcal{A}$, so by they are $C^k$ compatible with each other. **(3) $\mathcal{A}^{*}$ is maximal.** Suppose an $n$-dimensional chart $(U, \varphi)$ is $C^k$ compatible with every chart of $\mathcal{A}^{*}$. By (1) we have $\mathcal{A} \subset \mathcal{A}^{*}$, so in particular $(U,\varphi)$ is $C^k$ compatible with every chart of $\mathcal{A}$. This means $(U, \varphi) \in \mathcal{A}^{*}$. **(4) Uniqueness.** Let $\mathcal{B}$ be a maximal $C^k$ atlas with $\mathcal{A} \subset \mathcal{B}$. Each chart of $\mathcal{B}$ is $C^k$ compatible with all other charts of $\mathcal{B}$, in particular with all charts of $\mathcal{A} \subset \mathcal{B}$, so $\mathcal{B} \subset \mathcal{A}^{*}$. Conversely, let $(U, \varphi) \in \mathcal{A}^{*}$ and take an arbitrary $(V, \psi) \in \mathcal{B}$. Since $(V,\psi) \in \mathcal{B} \subset \mathcal{A}^{*}$, both $(U,\varphi)$ and $(V,\psi)$ are $C^k$ compatible with every chart of $\mathcal{A}$, so by they are $C^k$ compatible with each other. As $(V,\psi)$ was arbitrary, maximality of $\mathcal{B}$ gives $(U, \varphi) \in \mathcal{B}$. Hence $\mathcal{A}^{*} \subset \mathcal{B}$, and therefore $\mathcal{B} = \mathcal{A}^{*}$. Two practical consequences are worth recording. First, to construct a manifold **it suffices to write down one atlas**. By , the $C^k$ structure it determines is unique. All the examples below follow this practice. Second, if $(U, \varphi)$ belongs to a maximal atlas $\mathcal{A}$ and $\emptyset \ne U' \subset U$ is open, then $(U', \varphi|_{U'})$ also belongs to $\mathcal{A}$. The reason is as follows. The pair $(U', \varphi|_{U'})$ is a chart, and its transition map with any $(V, \psi) \in \mathcal{A}$ is the restriction to the open set $\varphi(U' \cap V)$ of the transition map between $(U,\varphi)$ and $(V,\psi)$, hence is $C^k$. Maximality then puts it in $\mathcal{A}$. In other words, **the domain of a chart may always be shrunk**, a fact we use in §6 when dealing with the $C^k$ property of maps. Note finally that if the definition is extended to $k = 0$, then any two charts of the same dimension are automatically $C^0$ compatible (transition maps are homeomorphisms, hence continuous). So a $C^0$ structure is determined by the topology and carries no new information. Differentiable structures acquire content only for $k \ge 1$. ### 3.4. The dimension is well defined In we fixed $n$ from the outset by speaking of a family of $n$-dimensional charts. Let us verify that this is not an unnatural restriction. Let $M$ be a topological space, $(U, \varphi)$ an $n$-dimensional chart of $M$ and $(V, \psi)$ an $m$-dimensional chart of $M$, with $U \cap V \ne \emptyset$. Assume further that the transition maps $\psi \circ \varphi^{-1} \colon \varphi(U \cap V) \to \psi(U \cap V)$ and $\varphi \circ \psi^{-1} \colon \psi(U \cap V) \to \varphi(U \cap V)$ are both of class $C^1$. Then $n = m$. Take $p \in U \cap V$ and put $a = \varphi(p) \in \mathbb{R}^n$ and $b = \psi(p) \in \mathbb{R}^m$. As observed in §3.2, the set $\varphi(U \cap V)$ is open in $\mathbb{R}^n$ and $\psi(U \cap V)$ is open in $\mathbb{R}^m$. Put $F = \psi \circ \varphi^{-1}$ and $G = \varphi \circ \psi^{-1}$. These are mutually inverse, so $$ G \circ F = \mathrm{id}_{\varphi(U \cap V)}, \qquad F \circ G = \mathrm{id}_{\psi(U \cap V)}. $$ By hypothesis $F$ and $G$ are both $C^1$, so applying part 2 of (the chain rule) to the first identity at the point $a$ gives $$ DG(F(a)) \, DF(a) = DG(b) \, DF(a) = I_n $$ (we used $F(a) = \psi(\varphi^{-1}(a)) = \psi(p) = b$). Similarly, applying it to the second identity at $b$ gives $DF(G(b)) \, DG(b) = DF(a)\, DG(b) = I_m$. Hence the linear maps $DF(a) \colon \mathbb{R}^n \to \mathbb{R}^m$ and $DG(b) \colon \mathbb{R}^m \to \mathbb{R}^n$ are mutually inverse linear isomorphisms. Isomorphic vector spaces have equal dimension (see [Vector spaces and linear transformations](/en/mathematics/linear-algebra/vector-spaces), in particular ), so $n = m$. Thanks to , on a space carrying an atlas of class $C^1$ or better, the dimension of a chart around each point is uniquely determined. The dimension is locally constant on $M$, so if $M$ is connected it is constant on all of $M$. For disconnected spaces some authors allow different dimensions on different components; in this article we fix $n$ from the start. Note that if no differentiability of the transition maps is assumed (the $C^0$ case), the same conclusion still holds, but its proof requires Brouwer's invariance of domain theorem, a deep result of topology. The gap between that and the $C^1$ case, which needs nothing beyond the chain rule, is wide: here we see the power of assuming a differentiable structure. ## 4. The definition of a differentiable manifold Let $k \in \mathbb{N} \cup \{\infty\}$ and let $n$ be an integer with $n \ge 0$. A pair $(M, \mathcal{A})$ is an **$n$-dimensional $C^k$ manifold** when 1. $M$ is a Hausdorff topological space, 2. $M$ is second countable, and 3. $\mathcal{A}$ is an $n$-dimensional differentiable structure of class $C^k$ on $M$ (that is, a maximal $n$-dimensional $C^k$ atlas). We call $n$ the **dimension** of $M$ and write $\dim M = n$. When $k = \infty$ we call $M$ a **smooth manifold** ($C^\infty$ manifold). When $\mathcal{A}$ is clear from the context we write simply $M$.
B["Assign an n-dimensional chart U, φ at each point"] B --> C["Atlas: the chart domains cover M"] C --> D["Ck compatibility: every two charts have Ck transition maps"] D --> E["Maximal atlas = Ck differentiable structure"] E --> F["n-dimensional Ck manifold"]`} />
Condition 3 was prepared in §3. By , it is in fact enough to give a single $n$-dimensional $C^k$ atlas to satisfy it. The issue is with conditions 1 and 2. These do not follow from the local condition (local Euclidean-ness), as the next example shows. Let $L$ be the quotient space obtained from the disjoint union of two lines $\mathbb{R} \times \{0\}$ and $\mathbb{R} \times \{1\}$ by identifying $(t, 0) \sim (t, 1)$ for $t \ne 0$. We write $[t, i]$ for equivalence classes. Thus $L$ is "a line with two origins". Put $U_i = \{[t, i] : t \in \mathbb{R}\}$ $(i = 0, 1)$ and define $\varphi_i([t,i]) = t$. Each $U_i$ is open (indeed $\pi^{-1}(U_i)$ is the union of $\mathbb{R} \times \{i\}$ and $(\mathbb{R}\setminus\{0\}) \times \{1-i\}$, which is open), and $\varphi_i$ is a homeomorphism. Since $U_0 \cup U_1 = L$, the family $\{(U_0, \varphi_0), (U_1, \varphi_1)\}$ is a $1$-dimensional atlas. The transition map is $$ \varphi_1 \circ \varphi_0^{-1} \colon \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}, \qquad t \mapsto t, $$ that is, the identity, which is $C^\infty$. So $L$ carries a $1$-dimensional $C^\infty$ atlas, and it is second countable as well (the images of countably many open intervals form a basis). However, $L$ is not Hausdorff. Consider the two points $0_0 = [0,0]$ and $0_1 = [0,1]$. Every open neighbourhood of $0_0$ contains $\varphi_0^{-1}((-\varepsilon, \varepsilon))$ for some $\varepsilon > 0$, and likewise every open neighbourhood of $0_1$ contains $\varphi_1^{-1}((-\delta, \delta))$ for some $\delta > 0$. Setting $\rho = \min(\varepsilon, \delta)/2 > 0$, the point $[\rho, 0] = [\rho, 1]$ belongs to both. So $0_0$ and $0_1$ cannot be separated by disjoint open sets. As a consequence, the sequence $[1/j, 0]$ $(j \in \mathbb{N})$, for instance, converges both to $0_0$ and to $0_1$. Once limits fail to be unique, basic arguments such as uniqueness of solutions of differential equations all collapse. This is why the Hausdorff condition is built into the definition. The reason for imposing second countability lies less in a pathology one can see immediately than in the tools one loses without it. A locally Euclidean space that is Hausdorff and second countable is paracompact, and therefore admits **partitions of unity**. Partitions of unity are the standard device for gluing locally defined objects (Riemannian metrics, volume elements, connections) into global ones, and they are also used to define integration on manifolds. Whitney's embedding theorem (an $n$-dimensional manifold embeds into $\mathbb{R}^{2n+1}$) likewise presupposes second countability. What happens without second countability is visible in a simple example. Take an uncountable set $A$ and give $X = \bigsqcup_{\alpha \in A} \mathbb{R}$ (disjoint union) the disjoint union topology; then $X$ is Hausdorff and carries a $1$-dimensional $C^\infty$ atlas, but is not second countable (see ). This example is disconnected; a connected counterexample is provided by the long line, which is Hausdorff and locally Euclidean but neither second countable nor metrizable. Some authors assume paracompactness in place of second countability. The only difference is that manifolds with uncountably many connected components are then allowed. **(a) $\mathbb{R}^n$ itself.** The single-chart family $\mathcal{A}_0 = \{(\mathbb{R}^n, \mathrm{id})\}$ is a $C^\infty$ atlas (compatibility need only be checked for $(U,\varphi)$ against itself, and the transition map is the identity, hence $C^\infty$). Since $\mathbb{R}^n$ is Hausdorff and second countable, the maximal atlas generated by $\mathcal{A}_0$ makes $\mathbb{R}^n$ an $n$-dimensional $C^\infty$ manifold. This is called the **standard differentiable structure** on $\mathbb{R}^n$. **(b) Open subsets of a manifold.** Let $(M, \mathcal{A})$ be an $n$-dimensional $C^k$ manifold and $W \subset M$ a non-empty open subset. Give $W$ the subspace topology; the Hausdorff condition and second countability are inherited by subspaces. Moreover $\mathcal{A}|_W = \{(U \cap W, \varphi|_{U \cap W}) : (U, \varphi) \in \mathcal{A},\ U \cap W \ne \emptyset\}$ is an $n$-dimensional $C^k$ atlas of $W$. Indeed, $U \cap W$ is open in $W$ and $\varphi|_{U\cap W}$ is a homeomorphism onto $\varphi(U \cap W)$; these sets cover $W$; and the transition maps are restrictions to open sets of the transition maps of $\mathcal{A}$, hence $C^k$. So by , $W$ is an $n$-dimensional $C^k$ manifold. It is called an **open submanifold**. **(c) The general linear group.** The set $M_n(\mathbb{R})$ of all real $n \times n$ matrices is identified with $\mathbb{R}^{n^2}$ by listing the entries, and is an $n^2$-dimensional $C^\infty$ manifold. The determinant $\det \colon M_n(\mathbb{R}) \to \mathbb{R}$ is a polynomial in the entries, hence continuous (see [Determinants and their properties](/en/mathematics/linear-algebra/determinants)), so $$ GL(n, \mathbb{R}) = \{A \in M_n(\mathbb{R}) : \det A \ne 0\} = {\det}^{-1}(\mathbb{R} \setminus \{0\}) $$ is open. By (b), therefore, $GL(n,\mathbb{R})$ is an $n^2$-dimensional $C^\infty$ manifold. This is the starting example for [Lie groups and Lie algebras](/mathematics/manifolds/lie-groups) (). ## 5. Examples: the sphere, real projective space and the torus From here on we construct explicit atlases on spaces that are not open subsets of $\mathbb{R}^n$. Throughout we take $k = \infty$, but the arguments are the same for any $k$. We begin with an operation that builds new manifolds from old. ### 5.1. Product manifolds Let $(M, \mathcal{A})$ be an $m$-dimensional $C^k$ manifold and $(N, \mathcal{B})$ an $n$-dimensional $C^k$ manifold. With the product topology, $M \times N$ becomes an $(m+n)$-dimensional $C^k$ manifold with the $C^k$ structure generated by the atlas $$ \mathcal{A} \times \mathcal{B} = \{ (U \times V,\ \varphi \times \psi) : (U, \varphi) \in \mathcal{A},\ (V, \psi) \in \mathcal{B} \}, \qquad (\varphi \times \psi)(p, q) = (\varphi(p), \psi(q)). $$ **The topological conditions.** A product of Hausdorff spaces is Hausdorff (if $(p,q) \ne (p',q')$ then the two differ in one factor, and we take the preimages of open sets separating that factor). A product of second countable spaces is second countable (if $\mathcal{U}$ and $\mathcal{V}$ are countable bases, then $\{B \times B' : B \in \mathcal{U}, B' \in \mathcal{V}\}$ is a countable basis). **These are charts.** The set $U \times V$ is open in the product topology, and $\varphi \times \psi$ is a bijection onto $\varphi(U) \times \psi(V) \subset \mathbb{R}^{m} \times \mathbb{R}^{n} = \mathbb{R}^{m+n}$; since $\varphi$ and $\psi$ are homeomorphisms, so is $\varphi \times \psi$. The set $\varphi(U) \times \psi(V)$ is a product of open sets, hence open in $\mathbb{R}^{m+n}$. These domains cover $M \times N$. **Compatibility.** Taking $(U', \varphi') \in \mathcal{A}$ and $(V', \psi') \in \mathcal{B}$, we have $(U \times V) \cap (U' \times V') = (U \cap U') \times (V \cap V')$, and the transition map is $$ (\varphi' \times \psi') \circ (\varphi \times \psi)^{-1}(u, v) = \big( (\varphi' \circ \varphi^{-1})(u),\ (\psi' \circ \psi^{-1})(v) \big). $$ The first component on the right is a $C^k$ function of $u$ alone and the second a $C^k$ function of $v$ alone, so as a function of $(u,v)$ all partial derivatives up to order $k$ exist and are continuous. Hence the map is $C^k$. Iterating this construction, a finite product of manifolds is a manifold. In particular, once we know that the circle $S^1$ is a manifold (next subsection), the $n$-dimensional torus $(S^1)^n$ is one too. ### 5.2. The sphere $S^n$ Give $S^n = \{x = (x^1, \ldots, x^{n+1}) \in \mathbb{R}^{n+1} : |x| = 1\}$ the subspace topology from $\mathbb{R}^{n+1}$. Since $\mathbb{R}^{n+1}$ is Hausdorff and second countable, so is $S^n$. Let $N = (0, \ldots, 0, 1)$ be the north pole and $S = (0, \ldots, 0, -1)$ the south pole, and set $$ U_N = S^n \setminus \{N\}, \quad \varphi_N(x) = \frac{(x^1, \ldots, x^n)}{1 - x^{n+1}}, \qquad U_S = S^n \setminus \{S\}, \quad \varphi_S(x) = \frac{(x^1, \ldots, x^n)}{1 + x^{n+1}}. $$ Geometrically $\varphi_N$ takes the intersection of the line through $N$ and $x$ with the hyperplane $x^{n+1} = 0$ (stereographic projection). **$\varphi_N$ is a homeomorphism.** For $x \in U_N$ we have $x^{n+1} \ne 1$, so the denominator never vanishes and $\varphi_N$ is continuous. As a candidate for the inverse take $$ \Phi(u) = \frac{(2u^1, \ldots, 2u^n,\ |u|^2 - 1)}{|u|^2 + 1} \qquad (u \in \mathbb{R}^n). $$ First, $\Phi(u) \in S^n$. Indeed $$ 4|u|^2 + (|u|^2 - 1)^2 = |u|^4 + 2|u|^2 + 1 = (|u|^2+1)^2, $$ so $|\Phi(u)|^2 = 1$. Also the $(n+1)$-st component is $(|u|^2-1)/(|u|^2+1) \ne 1$, so $\Phi(u) \in U_N$. Next we check $\varphi_N(\Phi(u)) = u$. Writing $\xi$ for the $(n+1)$-st component of $\Phi(u)$, we have $1 - \xi = 2/(|u|^2+1)$, hence $$ \varphi_N(\Phi(u)) = \frac{2u/(|u|^2+1)}{2/(|u|^2+1)} = u . $$ Conversely, for $x \in U_N$ put $u = \varphi_N(x)$. From $|x|=1$, $$ |u|^2 = \frac{1 - (x^{n+1})^2}{(1 - x^{n+1})^2} = \frac{1 + x^{n+1}}{1 - x^{n+1}}, \qquad |u|^2 + 1 = \frac{2}{1 - x^{n+1}}, \qquad |u|^2 - 1 = \frac{2 x^{n+1}}{1 - x^{n+1}}, $$ so the first $n$ components of $\Phi(u)$ are $2u \cdot (1-x^{n+1})/2 = (x^1, \ldots, x^n)$ and the $(n+1)$-st is $x^{n+1}$; that is, $\Phi(u) = x$. Since $\Phi$ is a rational expression whose denominator is at least $1$, it is continuous. Therefore $\varphi_N \colon U_N \to \mathbb{R}^n$ is a homeomorphism and $(U_N, \varphi_N)$ is an $n$-dimensional chart. The same computation works for $\varphi_S$ upon replacing $x^{n+1}$ by $-x^{n+1}$. **This is an atlas.** Since $N \ne S$, we have $U_N \cup U_S = S^n$. **The transition map.** We have $U_N \cap U_S = S^n \setminus \{N, S\}$ and $\varphi_N(U_N \cap U_S) = \mathbb{R}^n \setminus \{0\}$ (the equality $\varphi_N(x) = 0$ holds only for $x = S$). For $u \ne 0$, using the $(n+1)$-st component of $\Phi(u)$ we get $1 + \xi = 2|u|^2/(|u|^2+1)$, hence $$ \varphi_S \circ \varphi_N^{-1}(u) = \frac{2u/(|u|^2+1)}{2|u|^2/(|u|^2+1)} = \frac{u}{|u|^2}. $$ This is a $C^\infty$ map on $\mathbb{R}^n \setminus \{0\}$ (each component is a rational function whose denominator $|u|^2$ does not vanish). Moreover this map is its own inverse, so $\varphi_N \circ \varphi_S^{-1}$ is given by the same formula and is $C^\infty$ as well. Hence the two charts are $C^\infty$ compatible and $S^n$ is an $n$-dimensional $C^\infty$ manifold. **An atlas by graphs.** Another atlas can be built. For $i = 1, \ldots, n+1$ and a sign $\epsilon \in \{+1, -1\}$, set $$ V_i^{\epsilon} = \{x \in S^n : \epsilon x^i > 0\}, \qquad \psi_i^{\epsilon}(x) = (x^1, \ldots, \widehat{x^i}, \ldots, x^{n+1}) $$ (the hat indicates that the component is omitted). Then $\psi_i^{\epsilon}$ is a homeomorphism from $V_i^{\epsilon}$ onto the open ball $B^n = \{u \in \mathbb{R}^n : |u| < 1\}$, with inverse the map that inserts $\epsilon\sqrt{1 - |u|^2}$ in the $i$-th slot of $u$. Since some component of $x \in S^n$ is non-zero, these $2(n+1)$ charts cover $S^n$. A transition map inserts $\epsilon\sqrt{1-|u|^2}$ and deletes another component; as $\sqrt{1 - |u|^2}$ is $C^\infty$ on $B^n$ (the radicand is positive), the transition maps are $C^\infty$. Furthermore this atlas is $C^\infty$ compatible with the stereographic one. The map $\psi_i^{\epsilon} \circ \varphi_N^{-1}$ is $\Phi$ with its $i$-th component deleted, hence $C^\infty$. Conversely, $\varphi_N \circ (\psi_i^{\epsilon})^{-1}(u)$ forms $(x^1,\ldots,x^n)/(1 - x^{n+1})$ from the point $x = (\psi_i^\epsilon)^{-1}(u)$, whose components are $C^\infty$, and on the domain $x^{n+1} \ne 1$; so it is $C^\infty$. Hence by both atlases generate the same maximal atlas. That is, **either atlas yields the same differentiable structure**. ### 5.3. Real projective space $\mathbb{RP}^n$ On $\mathbb{R}^{n+1} \setminus \{0\}$ introduce the equivalence relation $x \sim y \iff \exists \lambda \in \mathbb{R} \setminus \{0\},\ y = \lambda x$, and consider the quotient space $\mathbb{RP}^n = (\mathbb{R}^{n+1}\setminus\{0\})/\!\sim$. Write $\pi$ for the quotient map and $\pi(x) = [x^0 : x^1 : \cdots : x^n]$ (**homogeneous coordinates**; indices run from $0$ to $n$). We identify $\mathbb{RP}^n$ with the set of all lines through the origin of $\mathbb{R}^{n+1}$. **$\pi$ is an open map.** For an open set $V \subset \mathbb{R}^{n+1}\setminus\{0\}$ we have $\pi^{-1}(\pi(V)) = \bigcup_{\lambda \ne 0} \lambda V$. For each $\lambda \ne 0$ the map $x \mapsto \lambda x$ is a homeomorphism, so $\lambda V$ is open, hence so is the union. By the definition of the quotient topology, $\pi(V)$ is open. **Second countability.** Let $\mathcal{U}$ be a countable basis of $\mathbb{R}^{n+1}\setminus\{0\}$. Then $\{\pi(B) : B \in \mathcal{U}\}$ is a countable family of open sets. Let us see that it is a basis. Let $W \subset \mathbb{RP}^n$ be open with $[x] \in W$; then $\pi^{-1}(W)$ is an open set containing $x$, so there is $B \in \mathcal{U}$ with $x \in B \subset \pi^{-1}(W)$. Then $[x] \in \pi(B) \subset \pi(\pi^{-1}(W)) = W$ (the last equality holds because $\pi$ is surjective). **The Hausdorff condition.** Put $R = \{(x,y) \in (\mathbb{R}^{n+1}\setminus\{0\})^2 : x \sim y\}$. For $x, y$ both non-zero, $x \sim y$ is equivalent to linear dependence of $x$ and $y$, which in turn is the vanishing of all $2 \times 2$ minors, that is, $$ x^i y^j - x^j y^i = 0 \qquad (0 \le i < j \le n). $$ The left-hand sides are continuous, so $R$ is an intersection of finitely many closed sets, hence closed. Here we invoke a general fact: if $\pi$ is open and $R$ is closed, then the quotient space is Hausdorff. Indeed, if $[x] \ne [y]$ then $(x,y) \notin R$, and since $R$ is closed there are open sets with $x \in A$, $y \in B$ and $(A \times B) \cap R = \emptyset$. As $\pi$ is open, $\pi(A)$ and $\pi(B)$ are open sets containing $[x]$ and $[y]$ respectively. If $\pi(A) \cap \pi(B) \ne \emptyset$, there would be $a \in A$ and $b \in B$ with $[a] = [b]$, so $(a, b) \in (A\times B) \cap R$, a contradiction. Hence $\pi(A) \cap \pi(B) = \emptyset$. **Charts.** For $i = 0, \ldots, n$ set $$ U_i = \{[x] \in \mathbb{RP}^n : x^i \ne 0\}, \qquad \varphi_i([x]) = \left( \frac{x^0}{x^i}, \ldots, \frac{x^{i-1}}{x^i}, \frac{x^{i+1}}{x^i}, \ldots, \frac{x^n}{x^i} \right) \in \mathbb{R}^n. $$ Both the condition $x^i \ne 0$ and the value of $\varphi_i$ are invariant under rescaling $x \mapsto \lambda x$, so neither depends on the representative. Since $\pi$ is open, $U_i = \pi(\{x : x^i \ne 0\})$ is open. Let us write down the inverse of $\varphi_i$. For $u = (u^1, \ldots, u^n) \in \mathbb{R}^n$, define the point of $\mathbb{R}^{n+1}$ $$ \iota_i(u) = (u^1, \ldots, u^i,\ 1,\ u^{i+1}, \ldots, u^n) $$ so that its $i$-th component (counting from $0$) equals $1$. Then $\varphi_i^{-1}(u) = [\iota_i(u)]$. Indeed, $\varphi_i([\iota_i(u)])$ divides by the $i$-th component $1$ of $\iota_i(u)$ and deletes that component, returning $u$; conversely if $x^i \ne 0$ then $[x] = [x/x^i] = [\iota_i(\varphi_i([x]))]$. The map $\varphi_i$ is continuous: $\varphi_i \circ \pi$ is a rational map on $\{x^i \ne 0\}$, hence continuous, and the restriction of $\pi$ to $\{x^i \ne 0\}$ is a quotient map onto $U_i$. And $\varphi_i^{-1} = \pi \circ \iota_i$ is continuous as a composite of continuous maps. Hence $(U_i, \varphi_i)$ is a chart, and since every $[x]$ has some $x^i \ne 0$, the family $\{(U_i, \varphi_i)\}_{i=0}^{n}$ covers $\mathbb{RP}^n$. **Transition maps.** The set $\varphi_i(U_i \cap U_j) = \{u \in \mathbb{R}^n : \iota_i(u)^j \ne 0\}$ is open, and on it $$ \varphi_j \circ \varphi_i^{-1}(u) = \left( \frac{\iota_i(u)^0}{\iota_i(u)^j}, \ldots, \widehat{\ \cdot\ }, \ldots, \frac{\iota_i(u)^n}{\iota_i(u)^j} \right) $$ (with the $j$-th component omitted). Each component is a quotient of affine functions of $u$ whose denominator does not vanish on the domain, hence is $C^\infty$. Therefore $\mathbb{RP}^n$ is an $n$-dimensional $C^\infty$ manifold. **A concrete case ($\mathbb{RP}^2$).** Since $\varphi_0([1 : u^1 : u^2]) = (u^1, u^2)$ and $\varphi_1([x^0:x^1:x^2]) = (x^0/x^1,\ x^2/x^1)$, on the region $u^1 \ne 0$ we get $$ \varphi_1 \circ \varphi_0^{-1}(u^1, u^2) = \varphi_1([1 : u^1 : u^2]) = \left( \frac{1}{u^1},\ \frac{u^2}{u^1} \right). $$ In the other direction $\varphi_0 \circ \varphi_1^{-1}(s, t) = \varphi_0([s : 1 : t]) = (1/s,\ t/s)$ for $s \ne 0$, again $C^\infty$. The space $\mathbb{RP}^n$ is also identified with the sphere $S^n$ with antipodal points identified. In physics it is important that $\mathbb{RP}^3$ is identified with the space of attitudes of a rigid body (the rotation group $SO(3)$). ### 5.4. The torus $T^n$ Let $\mathbb{Z}^n \subset \mathbb{R}^n$ be the integer lattice, $T^n = \mathbb{R}^n / \mathbb{Z}^n$ the quotient space and $\pi \colon \mathbb{R}^n \to T^n$ the quotient map ($\pi(x) = x + \mathbb{Z}^n$). **$\pi$ is an open map.** If $V$ is open then $\pi^{-1}(\pi(V)) = \bigcup_{m \in \mathbb{Z}^n}(V + m)$ is a union of open sets, hence open, so $\pi(V)$ is open. Second countability follows by the same argument as in . **The Hausdorff condition.** The map $d \colon \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$, $d(x,y) = y - x$, is continuous, and $\mathbb{Z}^n$ is closed in $\mathbb{R}^n$ (its points are isolated and it has no accumulation points). Hence the relation set $R = d^{-1}(\mathbb{Z}^n)$ is closed. As $\pi$ is open, the same argument used in shows that $T^n$ is Hausdorff. **Charts.** Let $V \subset \mathbb{R}^n$ be an open set satisfying "if $x, y \in V$ and $x - y \in \mathbb{Z}^n$ then $x = y$" (for example, any open cube of edge length less than $1$). Then $\pi|_V \colon V \to \pi(V)$ is a bijection which is continuous and open, hence a homeomorphism. So putting $$ \psi_V = (\pi|_V)^{-1} \colon \pi(V) \longrightarrow V \subset \mathbb{R}^n, $$ the pair $(\pi(V), \psi_V)$ is an $n$-dimensional chart. For any point $\pi(x)$ we may take the open cube of edge $1/2$ centred at $x$, so these charts cover $T^n$. **Transition maps.** Take two such sets $V, W$ and put $\Omega = \psi_V(\pi(V) \cap \pi(W)) \subset V$. Then $\Omega$ is open and the transition map is $\tau = \psi_W \circ \psi_V^{-1} = (\pi|_W)^{-1} \circ \pi|_{\Omega}$. For $x \in \Omega$ we have $\pi(\tau(x)) = \pi(x)$, hence $\tau(x) - x \in \mathbb{Z}^n$. The map $x \mapsto \tau(x) - x$ is continuous with values in the discrete set $\mathbb{Z}^n$, so the preimage of each $m \in \mathbb{Z}^n$ is an open and closed subset of $\Omega$; therefore this map is locally constant. That is, on each connected component of $\Omega$, $$ \tau(x) = x + m \qquad (m \in \mathbb{Z}^n \text{ constant}). $$ Translations are $C^\infty$, so the transition maps are $C^\infty$. Hence $T^n$ is an $n$-dimensional $C^\infty$ manifold. The torus $T^n$ is compact, since $\pi([0,1]^n) = T^n$ (see [Compactness](/mathematics/topology/compactness), in particular ). Moreover $T^n$ is diffeomorphic to $(S^1)^n$ (that $(S^1)^n$ is a manifold follows from $S^1 = T^1$ and ), and the configuration space $T^2$ of the double pendulum is this space. ## 6. $C^k$ functions and maps on a manifold The purpose of defining manifolds was to speak of differentiation on them. We begin with real-valued functions. Let $(M, \mathcal{A})$ be an $n$-dimensional $C^k$ manifold and $f \colon M \to \mathbb{R}$. We say $f$ is of class **$C^k$** when for every $p \in M$ there is a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$ whose **coordinate representation** $$ \hat{f} = f \circ \varphi^{-1} \colon \varphi(U) \longrightarrow \mathbb{R} $$ is of class $C^k$. We write $C^k(M)$ for the set of all $C^k$ functions on $M$. The definition has the form "there exists a chart", but in fact it is equivalent to "for every chart". This is the return on requiring compatibility. Let $(M, \mathcal{A})$ be an $n$-dimensional $C^k$ manifold and $f \colon M \to \mathbb{R}$. The following are equivalent. 1. $f$ is $C^k$ in the sense of . 2. For every chart $(V, \psi) \in \mathcal{A}$, the map $f \circ \psi^{-1}$ is $C^k$ on $\psi(V)$. That 2 implies 1 is clear (for each point take any chart containing it). We show 1 implies 2. Take an arbitrary $(V, \psi) \in \mathcal{A}$. By part 1 (locality) of , it suffices to show that $f \circ \psi^{-1}$ is $C^k$ on a neighbourhood of each point of $\psi(V)$. Take $s \in \psi(V)$ and put $p = \psi^{-1}(s) \in V$. By hypothesis 1 there is a chart $(U, \varphi) \in \mathcal{A}$ with $p \in U$ such that $f \circ \varphi^{-1}$ is $C^k$ on $\varphi(U)$. The set $\psi(U \cap V)$ is an open subset of $\mathbb{R}^n$ containing $s$, and on it $$ f \circ \psi^{-1} = (f \circ \varphi^{-1}) \circ (\varphi \circ \psi^{-1}). $$ Indeed, for $t \in \psi(U\cap V)$ we have $\psi^{-1}(t) \in U \cap V$, so $\varphi \circ \psi^{-1}(t) \in \varphi(U \cap V) \subset \varphi(U)$, and the value of the right-hand side is $f(\psi^{-1}(t))$. Since $(U,\varphi)$ and $(V,\psi)$ both belong to the maximal atlas $\mathcal{A}$, they are $C^k$ compatible, so $\varphi \circ \psi^{-1}$ is $C^k$ on $\psi(U \cap V)$. By hypothesis $f \circ \varphi^{-1}$ is $C^k$ on $\varphi(U) \supset \varphi(U \cap V)$. Hence part 2 (composition) of gives that $f \circ \psi^{-1}$ is $C^k$ on $\psi(U \cap V)$. As $s$ was arbitrary, locality gives that $f\circ\psi^{-1}$ is $C^k$ on $\psi(V)$. Let $(M, \mathcal{A})$ be an $m$-dimensional $C^k$ manifold, $(N, \mathcal{B})$ an $n$-dimensional $C^k$ manifold, and $F \colon M \to N$. 1. $F$ is of class **$C^k$** when for every $p \in M$ there are charts $(U, \varphi) \in \mathcal{A}$ with $p \in U$ and $(V, \psi) \in \mathcal{B}$ with $F(p) \in V$ satisfying $F(U) \subset V$, such that the coordinate representation $$ \psi \circ F \circ \varphi^{-1} \colon \varphi(U) \longrightarrow \psi(V) \subset \mathbb{R}^n $$ is of class $C^k$. 2. $F$ is a **$C^k$ diffeomorphism** when $F$ is a bijection and both $F$ and $F^{-1}$ are $C^k$. When a $C^k$ diffeomorphism from $M$ to $N$ exists, $M$ and $N$ are called **diffeomorphic**. Taking $N = \mathbb{R}$ with the chart $(\mathbb{R}, \mathrm{id})$, this definition agrees with . Note that continuity is not assumed in the definition; it follows automatically. If $F \colon M \to N$ is $C^k$ in the sense of , then $F$ is continuous. Take $p \in M$ and the charts $(U, \varphi)$, $(V, \psi)$ appearing in the definition. On $U$ we have $$ F|_U = \psi^{-1} \circ (\psi \circ F \circ \varphi^{-1}) \circ \varphi $$ (for $q \in U$ the right-hand side is $\psi^{-1}(\psi(F(q))) = F(q)$). Now $\varphi$ is a homeomorphism, hence continuous; the middle map is $C^k$, hence continuous; and $\psi^{-1}$ is a homeomorphism, hence continuous. So $F|_U$ is continuous. Let $O \subset N$ be open. For the set $U$ associated with each $p$ as above, $(F|_U)^{-1}(O) = F^{-1}(O) \cap U$ is open in $U$, hence open in $M$. As $p$ varies these sets $U$ cover $M$, so $$ F^{-1}(O) = \bigcup_{p \in M} \big( F^{-1}(O) \cap U_p \big) $$ is open, being a union of open sets. Therefore $F$ is continuous. Let $M$, $N$, $P$ be $C^k$ manifolds of dimensions $m$, $n$, $l$, with differentiable structures $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ respectively. 1. (Independence of charts) If $F \colon M \to N$ is $C^k$, then for any charts $(U, \varphi) \in \mathcal{A}$ and $(V, \psi) \in \mathcal{B}$ with $F(U) \subset V$, the map $\psi \circ F \circ \varphi^{-1}$ is $C^k$ on $\varphi(U)$. 2. (Composition) If $F \colon M \to N$ and $G \colon N \to P$ are both $C^k$, then $G \circ F \colon M \to P$ is $C^k$. **Proof of 1.** Take $(U,\varphi) \in \mathcal{A}$ and $(V,\psi) \in \mathcal{B}$ with $F(U) \subset V$, take an arbitrary $a \in \varphi(U)$ and put $p = \varphi^{-1}(a)$. By part 1 of , it suffices to prove that the map is $C^k$ on a neighbourhood of $a$. Since $F$ is $C^k$, there are charts $(U', \varphi') \in \mathcal{A}$ and $(V', \psi') \in \mathcal{B}$ with $p \in U'$, $F(p) \in V'$, $F(U') \subset V'$, such that $\psi' \circ F \circ \varphi'^{-1}$ is $C^k$. The set $W = U \cap U'$ is open and contains $p$, and for $q \in W$ we have $F(q) \in V$ and $F(q) \in V'$, that is, $F(W) \subset V \cap V'$. The set $\varphi(W)$ is open and contains $a$, and on it $$ \psi \circ F \circ \varphi^{-1} = (\psi \circ \psi'^{-1}) \circ (\psi' \circ F \circ \varphi'^{-1}) \circ (\varphi' \circ \varphi^{-1}). $$ Indeed, for $s \in \varphi(W)$ put $q = \varphi^{-1}(s) \in W$; the rightmost map sends $s$ to $\varphi'(q) \in \varphi'(W) \subset \varphi'(U')$, the middle one sends this to $\psi'(F(q)) \in \psi'(V \cap V')$, and the leftmost one sends that to $\psi(F(q))$. All three maps are $C^k$: the map $\varphi' \circ \varphi^{-1}$ because $(U,\varphi)$ and $(U',\varphi')$ both belong to the maximal atlas $\mathcal{A}$, the map $\psi \circ \psi'^{-1}$ likewise by compatibility within $\mathcal{B}$, and the middle one by hypothesis. Hence part 2 of makes the composite $C^k$ on $\varphi(W)$. As $a$ was arbitrary, the claim follows. **Proof of 2.** Take $p \in M$. Since $G$ is $C^k$, there are charts $(V_0, \psi_0) \in \mathcal{B}$ and $(W_0, \chi) \in \mathcal{C}$ with $F(p) \in V_0$, $G(F(p)) \in W_0$ and $G(V_0) \subset W_0$ such that $\chi \circ G \circ \psi_0^{-1}$ is $C^k$. Next we choose a chart on the $M$ side. By the map $F$ is continuous, so $F^{-1}(V_0)$ is an open set containing $p$. Take any chart $(U_1, \varphi_1) \in \mathcal{A}$ containing $p$ and put $U = U_1 \cap F^{-1}(V_0)$, $\varphi = \varphi_1|_U$; by we have $(U, \varphi) \in \mathcal{A}$, and $F(U) \subset V_0$. Applying part 1 to $(U,\varphi)$ and $(V_0, \psi_0)$, the map $\psi_0 \circ F \circ \varphi^{-1}$ is $C^k$ on $\varphi(U)$. This map sends $\varphi(U)$ into $\psi_0(V_0)$, and $\chi \circ G \circ \psi_0^{-1}$ is $C^k$ on $\psi_0(V_0)$, so by part 2 of the map $$ \chi \circ (G \circ F) \circ \varphi^{-1} = (\chi \circ G \circ \psi_0^{-1}) \circ (\psi_0 \circ F \circ \varphi^{-1}) $$ is $C^k$ on $\varphi(U)$. Also $(G\circ F)(U) \subset G(V_0) \subset W_0$, so $(U,\varphi)$ and $(W_0, \chi)$ satisfy the condition in . As $p$ was arbitrary, $G \circ F$ is $C^k$. **(a) The inclusion $\iota \colon S^n \hookrightarrow \mathbb{R}^{n+1}$.** Take $(\mathbb{R}^{n+1}, \mathrm{id})$ as a chart of $\mathbb{R}^{n+1}$ and the stereographic chart $(U_N, \varphi_N)$ of $S^n$. The condition $\iota(U_N) \subset \mathbb{R}^{n+1}$ holds trivially, and the coordinate representation is exactly the map $\Phi$ of , namely $$ \mathrm{id} \circ \iota \circ \varphi_N^{-1}(u) = \frac{(2u^1, \ldots, 2u^n,\ |u|^2-1)}{|u|^2+1}. $$ The denominator is at least $1$, so each component is a $C^\infty$ rational function on $\mathbb{R}^n$. The same holds for $(U_S, \varphi_S)$, so $\iota$ is $C^\infty$. **(b) The double cover $\rho \colon S^n \to \mathbb{RP}^n$.** Define $\rho(x) = [x]$ (this makes sense since $x \ne 0$). For a point $x \in S^n$ choose $i$ with $x^i \ne 0$, and take the graph chart $(V_i^{\epsilon}, \psi_i^{\epsilon})$ of $S^n$ (with $\epsilon$ the sign of $x^i$; here indices run from $0$ to $n$) together with the chart $(U_i, \varphi_i)$ of $\mathbb{RP}^n$. If $y \in V_i^\epsilon$ then $y^i \ne 0$, so $\rho(V_i^\epsilon) \subset U_i$. For $u \in B^n$, the $i$-th component of $y = (\psi_i^\epsilon)^{-1}(u)$ is $\epsilon\sqrt{1-|u|^2}$ and the remaining components are those of $u$, so $$ \varphi_i \circ \rho \circ (\psi_i^{\epsilon})^{-1}(u) = \frac{u}{\epsilon\sqrt{1 - |u|^2}}. $$ The denominator does not vanish on $B^n$ and is $C^\infty$ there, so this map is $C^\infty$. Hence $\rho$ is $C^\infty$. **(c) The quotient map $\pi \colon \mathbb{R}^n \to T^n$.** Taking the chart $(\pi(V), \psi_V)$ of , we have $\pi(V) \subset \pi(V)$ and $$ \psi_V \circ \pi \circ \mathrm{id}^{-1}|_V = (\pi|_V)^{-1} \circ \pi|_V = \mathrm{id}_V, $$ so the coordinate representation is the identity, which is $C^\infty$. Every point of $\mathbb{R}^n$ lies in such a $V$, so $\pi$ is $C^\infty$. Moreover, since the coordinate representation is the identity, $\pi$ maps a neighbourhood of each point $C^\infty$-diffeomorphically onto a neighbourhood of its image (it is a local diffeomorphism). One topological space can carry different differentiable structures. On $\mathbb{R}$ put $h(t) = t^3$; since $h$ is a homeomorphism, both $\mathcal{A}_1 = \{(\mathbb{R}, \mathrm{id})\}$ and $\mathcal{A}_2 = \{(\mathbb{R}, h)\}$ are $1$-dimensional $C^\infty$ atlases. Yet they are not $C^1$ compatible: one transition map $h \circ \mathrm{id}^{-1} = h$ is $C^\infty$, but the other, $\mathrm{id} \circ h^{-1}(s) = s^{1/3}$, is not differentiable at $s = 0$. So the maximal atlases generated by $\mathcal{A}_1$ and $\mathcal{A}_2$ are different. Even so, the two manifolds are diffeomorphic. Define $F \colon (\mathbb{R}, \mathcal{A}_2^{*}) \to (\mathbb{R}, \mathcal{A}_1^{*})$ by $F(t) = t^3$. Its coordinate representation is $\mathrm{id} \circ F \circ h^{-1}(s) = (s^{1/3})^3 = s$, the identity, which is $C^\infty$; and the coordinate representation of the inverse $F^{-1}(s) = s^{1/3}$ is $h \circ F^{-1} \circ \mathrm{id}^{-1}(t) = (t^{1/3})^3 = t$, again $C^\infty$. In other words, "the differentiable structures differ" and "the manifolds differ" are two distinct assertions. Whether there exist manifolds that are homeomorphic but not diffeomorphic is a deep question. In 1956 Milnor constructed $7$-dimensional manifolds homeomorphic but not diffeomorphic to $S^7$ (exotic spheres). Moreover, by results in $4$-dimensional topology from the 1980s, it is known that for $n \ne 4$ the differentiable structure on $\mathbb{R}^n$ is unique up to the standard one, whereas $\mathbb{R}^4$ carries uncountably many non-standard differentiable structures. ## 7. Exercises Let $A$ be an uncountable set and give $X = \bigsqcup_{\alpha \in A} \mathbb{R}_\alpha$ (where each $\mathbb{R}_\alpha$ is a copy of $\mathbb{R}$) the disjoint union topology: a set $W \subset X$ is open when $W \cap \mathbb{R}_\alpha$ is open in $\mathbb{R}_\alpha$ for every $\alpha$. Show that $X$ is Hausdorff and carries a $1$-dimensional $C^\infty$ atlas, but is not second countable. **The Hausdorff condition.** Take two distinct points $p, q \in X$. If they lie in the same $\mathbb{R}_\alpha$, then since $\mathbb{R}$ is Hausdorff they can be separated by disjoint open sets inside $\mathbb{R}_\alpha$, and those are open in $X$. If they lie in different copies $\mathbb{R}_\alpha$ and $\mathbb{R}_\beta$, then $\mathbb{R}_\alpha$ and $\mathbb{R}_\beta$ themselves are disjoint open sets. **The atlas.** Put $U_\alpha = \mathbb{R}_\alpha$ and let $\varphi_\alpha \colon U_\alpha \to \mathbb{R}$ be the natural identification. Each $(U_\alpha, \varphi_\alpha)$ is a $1$-dimensional chart and $\bigcup_\alpha U_\alpha = X$. For $\alpha \ne \beta$ we have $U_\alpha \cap U_\beta = \emptyset$, so compatibility holds automatically, and for $\alpha = \beta$ the transition map is the identity, which is $C^\infty$. Hence $\{(U_\alpha, \varphi_\alpha)\}$ is a $1$-dimensional $C^\infty$ atlas. **Failure of second countability.** Let $\mathcal{U}$ be a basis for the topology of $X$. For each $\alpha \in A$ choose a point $p_\alpha \in \mathbb{R}_\alpha$. Since $\mathbb{R}_\alpha$ is an open set containing $p_\alpha$, the definition of a basis gives some $B_\alpha \in \mathcal{U}$ with $p_\alpha \in B_\alpha \subset \mathbb{R}_\alpha$. For $\alpha \ne \beta$ the sets $B_\alpha \subset \mathbb{R}_\alpha$ and $B_\beta \subset \mathbb{R}_\beta$ are disjoint and both non-empty, so $B_\alpha \ne B_\beta$. Hence $\alpha \mapsto B_\alpha$ is injective and the cardinality of $\mathcal{U}$ is at least that of $A$, hence uncountable. Therefore no countable basis exists. (This $X$ fails condition 2 of , so it is not a manifold.) Let $M$ and $N$ be $C^k$ manifolds and give $M \times N$ the $C^k$ structure of . Show the following. 1. The projections $\mathrm{pr}_M \colon M \times N \to M$ and $\mathrm{pr}_N \colon M \times N \to N$ are both $C^k$. 2. Let $P$ be a $C^k$ manifold and $H \colon P \to M \times N$. Then $H$ is $C^k$ if and only if $\mathrm{pr}_M \circ H$ and $\mathrm{pr}_N \circ H$ are both $C^k$. **1.** Take $(p, q) \in M \times N$ and charts $(U,\varphi) \in \mathcal{A}$, $(V,\psi) \in \mathcal{B}$ with $p \in U$ and $q \in V$. For the product chart $(U \times V, \varphi \times \psi)$ we have $\mathrm{pr}_M(U \times V) = U \subset U$, and the coordinate representation is $$ \varphi \circ \mathrm{pr}_M \circ (\varphi \times \psi)^{-1}(u, v) = \varphi(\varphi^{-1}(u)) = u, $$ that is, the projection $\mathbb{R}^{m+n} \supset \varphi(U)\times\psi(V) \to \mathbb{R}^m$ onto the first factor. Being linear, it is $C^\infty$ and in particular $C^k$. The same applies to $\mathrm{pr}_N$. **2.** If $H$ is $C^k$, then by 1 and part 2 (composition) of , the maps $\mathrm{pr}_M \circ H$ and $\mathrm{pr}_N \circ H$ are $C^k$. Conversely, assume $H_1 = \mathrm{pr}_M \circ H$ and $H_2 = \mathrm{pr}_N \circ H$ are both $C^k$. Take $r \in P$ and charts $(U,\varphi)$, $(V,\psi)$ with $H_1(r) \in U$ and $H_2(r) \in V$. By the maps $H_1$, $H_2$ are continuous, so $H_1^{-1}(U) \cap H_2^{-1}(V)$ is an open set containing $r$. Let $(W, \chi)$ be a chart of $P$ containing $r$, restricted to this open set (by this is again a chart). Then $H_1(W) \subset U$ and $H_2(W) \subset V$, that is, $H(W) \subset U \times V$. The coordinate representation is $$ (\varphi \times \psi) \circ H \circ \chi^{-1}(w) = \big( \varphi \circ H_1 \circ \chi^{-1}(w),\ \psi \circ H_2 \circ \chi^{-1}(w) \big), $$ and by part 1 of each component is $C^k$ on $\chi(W)$. A vector-valued map is $C^k$ exactly when each of its components is, so the whole map is $C^k$. As $r$ was arbitrary, $H$ is $C^k$. Show that $\mathbb{RP}^1$ and $S^1$ are diffeomorphic. Hint: identify $\mathbb{R}^2$ with $\mathbb{C}$ and consider $F([z]) = z^2/|z|^2$. Identify $\mathbb{R}^2 \ni (x^0, x^1) \leftrightarrow z = x^0 + i x^1 \in \mathbb{C}$ and define $F \colon \mathbb{RP}^1 \to S^1$ by $F([z]) = z^2/|z|^2$. **Well defined.** For $\lambda \in \mathbb{R} \setminus \{0\}$ we have $(\lambda z)^2/|\lambda z|^2 = \lambda^2 z^2/(\lambda^2 |z|^2) = z^2/|z|^2$, so the value does not depend on the representative. Also $|z^2/|z|^2| = 1$, so the value lies in $S^1$. **Bijectivity.** We may normalise $|z| = 1$, and then $F([z]) = z^2$. For any $w \in S^1$ there are exactly two $z$ with $|z|=1$ and $z^2 = w$, namely $z$ and $-z$, which form a single equivalence class. Hence $F$ is a bijection. **Smoothness.** Take the chart $\varphi_0([1 : u]) = u$ of $\mathbb{RP}^1$, so that $z = 1 + iu$, and $$ F(\varphi_0^{-1}(u)) = \frac{(1+iu)^2}{1+u^2} = \frac{1 - u^2}{1+u^2} + i\,\frac{2u}{1+u^2}, $$ that is, the point $(a, b) = \big((1-u^2)/(1+u^2),\ 2u/(1+u^2)\big)$ of $S^1$. On the $S^1$ side we use the stereographic projection from the north pole $(0,1)$, namely $\varphi_N(a,b) = a/(1-b)$. We have $F(\varphi_0^{-1}(u)) = (0,1)$ only when $1 - u^2 = 0$ and $2u = 1 + u^2$, that is, only for $u = 1$, so we restrict the domain to $u \ne 1$. There $$ 1 - b = 1 - \frac{2u}{1+u^2} = \frac{(1-u)^2}{1+u^2}, \qquad \varphi_N \circ F \circ \varphi_0^{-1}(u) = \frac{1-u^2}{(1-u)^2} = \frac{1+u}{1-u}, $$ which is $C^\infty$ for $u \ne 1$. Near $u = 1$ we replace the chart on the $S^1$ side by the stereographic projection from the south pole, $\varphi_S(a,b) = a/(1+b)$. Since $1 + b = (1+u)^2/(1+u^2)$, we get $\varphi_S \circ F \circ \varphi_0^{-1}(u) = (1-u)/(1+u)$, which is $C^\infty$ for $u \ne -1$. As $u = -1$ is not near $u = 1$, these two charts cover the domain of $\varphi_0$. Near the remaining point $[0:1]$ we use the chart $\varphi_1([s : 1]) = s$ and carry out the same computation with $z = s + i$; again a rational expression results, so the map is $C^\infty$. Hence $F$ is $C^\infty$. **Smoothness of the inverse.** The expression $g(u) = (1+u)/(1-u)$ obtained above is a bijection from $u \ne 1$ onto $s \ne -1$, with inverse $g^{-1}(s) = (s-1)/(s+1)$. This is $C^\infty$ for $s \ne -1$, and it is exactly the coordinate representation $\varphi_0 \circ F^{-1} \circ \varphi_N^{-1}$. The other pairs of charts give rational expressions in the same way, with denominators that do not vanish on the respective domains. Hence $F^{-1}$ is also $C^\infty$, and $F$ is a diffeomorphism. Let $A \in GL(n, \mathbb{R})$ and consider the quotient $\mathbb{R}^n/\Lambda$ by the lattice $\Lambda = A\mathbb{Z}^n = \{Am : m \in \mathbb{Z}^n\}$. Show that $\mathbb{R}^n/\Lambda$ is an $n$-dimensional $C^\infty$ manifold and that it is diffeomorphic to $T^n = \mathbb{R}^n/\mathbb{Z}^n$. **It is a manifold.** Since $A$ is a linear homeomorphism, $\Lambda = A\mathbb{Z}^n$ is a discrete closed subgroup of $\mathbb{R}^n$. In particular $\delta = \min\{|v| : v \in \Lambda,\ v \ne 0\}$ exists and $\delta > 0$ (as $\Lambda$ is discrete and closed, only finitely many of its points lie in a bounded closed ball centred at the origin). Let $\pi_\Lambda \colon \mathbb{R}^n \to \mathbb{R}^n/\Lambda$ be the quotient map. By the same arguments as in , the map $\pi_\Lambda$ is open, $\mathbb{R}^n/\Lambda$ is second countable, and it is Hausdorff because $R = d^{-1}(\Lambda)$ (with $d(x,y) = y-x$) is closed. Taking an open set $V$ of diameter less than $\delta$, the restriction $\pi_\Lambda|_V$ is a homeomorphism, so $(\pi_\Lambda(V), (\pi_\Lambda|_V)^{-1})$ is a chart, and the transition maps are translations by elements of $\Lambda$ (locally constant), hence $C^\infty$. Therefore $\mathbb{R}^n/\Lambda$ is an $n$-dimensional $C^\infty$ manifold. **They are diffeomorphic.** Since $x - y \in \mathbb{Z}^n \iff Ax - Ay \in \Lambda$, the map $\bar{A}(\pi(x)) = \pi_\Lambda(Ax)$ is well defined. As $A$ is bijective and $A\mathbb{Z}^n = \Lambda$, the map $\bar{A}$ is bijective, with inverse the map $\overline{A^{-1}}$ induced by $A^{-1}$. Let us check smoothness. Choose $V \subset \mathbb{R}^n$ small enough to give a chart of $T^n$ and such that $A(V)$ has diameter less than $\delta$ (such a $V$ exists around each point since $A$ is continuous). Then $\bar{A}(\pi(V)) = \pi_\Lambda(A(V))$, and the coordinate representation is $$ (\pi_\Lambda|_{A(V)})^{-1} \circ \bar{A} \circ \pi|_V = A|_V . $$ Indeed, for $x \in V$ the left-hand side takes $\pi_\Lambda(Ax)$ back to its representative in $A(V)$, and since $Ax \in A(V)$ that value is $Ax$. The linear map $A$ is $C^\infty$, so $\bar{A}$ is $C^\infty$. Applying the same argument to $A^{-1}$ shows that $\overline{A^{-1}} = \bar{A}^{-1}$ is $C^\infty$. Hence $\bar{A}$ is a diffeomorphism. (On the other hand, if $\mathbb{R}^n/\Lambda$ is given the metric induced by the standard inner product of $\mathbb{R}^n$, differences in the shape of the lattice persist as differences of lengths and angles. That two spaces may be diffeomorphic yet metrically different is where Riemannian geometry begins.) ## References - J. M. Lee, *Introduction to Smooth Manifolds*, 2nd edition, Springer, 2013 — Chapter 1 "Smooth Manifolds", Chapter 2 "Smooth Maps". Charts, atlases and maximal atlases are treated exactly as in this article. - Y. Matsushima, *Tayōtai Nyūmon* (in Japanese), Shōkabō, 1965 — the chapter on the definition of a differentiable manifold and atlases. A standard Japanese introduction. - Y. Matsumoto, *Tayōtai no Kiso* (in Japanese), University of Tokyo Press, 1988 — the chapter on the definition of a manifold and concrete examples, written carefully from the topological preliminaries onwards. - B. Riemann, "Über die Hypothesen, welche der Geometrie zu Grunde liegen", *Abhandlungen der Königlichen Gesellschaft der Wissenschaften zu Göttingen* 13 (1868) — the 1854 inaugural lecture in which the notion of a manifold was first put forward. - L. A. Steen and J. A. Seebach, Jr., *Counterexamples in Topology*, 2nd edition, Springer, 1978 — counterexamples such as the line with two origins and the long line are collected here. - J. Milnor, "On manifolds homeomorphic to the 7-sphere", *Annals of Mathematics* 64 (1956), 399–405 — the first construction of exotic spheres. ## Appendix: Assembling a manifold from charts **Putting the topology in afterwards.** In this article we built manifolds by first providing a topological space $M$ and then placing charts on it. In practice, however, fixing the topology first can be laborious. For $\mathbb{RP}^n$, for instance, we had to bring in $2 \times 2$ minors just to verify that the quotient topology is Hausdorff. In such cases the order can be reversed. Suppose we are given a set $M$, a family of subsets $\{U_\alpha\}_{\alpha \in A}$ and injections $\varphi_\alpha \colon U_\alpha \to \mathbb{R}^n$ satisfying the following five conditions. (i) Each $\varphi_\alpha(U_\alpha)$ is open in $\mathbb{R}^n$. (ii) For all $\alpha, \beta$, the set $\varphi_\alpha(U_\alpha \cap U_\beta)$ is open in $\mathbb{R}^n$. (iii) Whenever $U_\alpha \cap U_\beta \ne \emptyset$, the map $\varphi_\beta \circ \varphi_\alpha^{-1} \colon \varphi_\alpha(U_\alpha \cap U_\beta) \to \varphi_\beta(U_\alpha \cap U_\beta)$ is $C^k$. (iv) Countably many of the $U_\alpha$ cover $M$. (v) For any two distinct points $p, q$ of $M$, either some $U_\alpha$ contains both, or there are $\alpha, \beta$ with $p \in U_\alpha$, $q \in U_\beta$ and $U_\alpha \cap U_\beta = \emptyset$. Then there is exactly one topology together with a $C^k$ structure on $M$ making each $(U_\alpha, \varphi_\alpha)$ a chart. The topology is defined by taking $\{\varphi_\alpha^{-1}(B) : \alpha \in A,\ B \subset \varphi_\alpha(U_\alpha)$ an open ball$\}$ as a basis; condition (iv) guarantees second countability and (v) the Hausdorff condition. A proof is in Lee's textbook (the Smooth Manifold Chart Lemma in Chapter 1). **Why this viewpoint pays.** In this form, building a manifold reduces uniformly to "lay out charts and check that the transition maps are $C^k$". The spaces $S^n$, $\mathbb{RP}^n$ and $T^n$ of §5 can all be rebuilt by this procedure. Moreover, for manifolds with boundary as treated in [Stokes' theorem (general form)](/mathematics/manifolds/stokes-theorem), and for total spaces of fibre bundles and other spaces that are naturally defined by gluing, this construction is in practice the only route available.