Smooth Manifolds and Smooth Maps

Let $n\in \mathbb{N}$. For $n\ge 1$, put ${\mathbb{R}}^n:={\prod }_{k<n}\mathbb{R}$ with its usual topology; for $n=0$ put ${\mathbb{R}}^0:=\{0\}$, the one-point space. A topological $n$-manifold without boundary (or briefly an $n$-manifold) is a topological space $M$ satisfying:

1. $M$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not);
2. $M$ is second countable (Second countability: an at most countable basis for the topology);
3. $M$ is locally Euclidean of dimension $n$: every $p\in M$ has an open neighbourhood $U\subseteq M$ homeomorphic to an open subset of ${\mathbb{R}}^n$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

The empty space satisfies all three conditions vacuously, so $M=\mathrm{\varnothing }$ is an $n$-manifold for every $n$; this degenerate instance is kept, and statements about nonempty manifolds name the hypothesis. In dimension zero, condition 3 forces the one-point neighbourhoods of points to be open singletons, so a $0$-manifold is exactly a discrete second-countable space with at most countably many points.

• Second countability is part of the definition. It is exactly the axiom that the long line fails, and it is what later turns the local hypotheses into global ones: countable chart bases, $\sigma$-compactness, metrizability and paracompactness all flow from it. The convention split is recorded in Manifold conventions and the role of second countability.

• No boundary is defined here. A manifold with boundary replaces the local models by open subsets of the closed upper half-space; that is a strictly later construction and no statement on this page silently permits it.

Let $M$ be a topological $n$-manifold (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces). A chart on $M$ is a pair $(U,\phi )$ in which:

• $U\subseteq M$ is open, called the coordinate domain;
• $\phi :U\to \widehat{U}$ is a homeomorphism onto an open subset $\widehat{U}\subseteq {\mathbb{R}}^n$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), called the coordinate map; the open set $\widehat{U}=\phi (U)$ is the chart image.

For $n\ge 1$ and $i<n$, the $i$-th coordinate function is $x^i:={\pi }_i\circ \phi :U⟶\mathbb{R},x^i(p)=\phi (p{)}_i,$ so that $\phi =(x^0,\dots ,x^{n-1})$. A chart is often written $(U,(x^0,\dots ,x^{n-1}))$ and the coordinates are then used to name points of $U$. In dimension zero the coordinate map is the unique map to $\{0\}$ and there are no coordinate functions.

• The domain is open in the manifold, the image is open in Euclidean space. The two openness statements are separate hypotheses; in particular the domain $U$ need not itself be an open subset of ${\mathbb{R}}^n$, although the chart map makes it homeomorphic to the Euclidean open set $\widehat{U}$. The companion false statement A chart domain need not be a Euclidean open set records the failure to keep them apart.

• A chart is a homeomorphism by definition, so $\phi$ is continuous, bijective, and ${\phi }^{-1}:\widehat{U}\to U$ is continuous. Smoothness of either map is a later condition on pairs of charts, not a hypothesis here.

Let $M$ be a topological $m$-manifold, let $N$ be a topological $n$-manifold, let $F:M\to N$ be any function, and let $(U,\phi )$ and $(V,\psi )$ be charts on $M$ and $N$ respectively (Manifold charts, coordinate domains, and coordinate functions). The coordinate representation (or local representative) of $F$ with respect to these charts is the function

$\widehat{F}:=\psi \circ F\circ {\phi }^{-1}:\phi (U\cap F^{-1}(V))⟶\psi (V),$

which is defined on the image of $U\cap F^{-1}(V)$. When $F$ is continuous, the preimage $F^{-1}(V)$ is open in $M$, so $U\cap F^{-1}(V)$ is open in the open set $U$ and its image under the homeomorphism $\phi$ is an open subset of ${\mathbb{R}}^m$; in that case $\widehat{F}$ is a map between open subsets of Euclidean spaces, and only such representatives are ever tested for smoothness.

With coordinates $(x^1,\dots ,x^m)$ on $U$ and $(y^1,\dots ,y^n)$ on $V$, the representative expresses the image coordinates as functions of the source coordinates: $\widehat{F}(x^1,\dots ,x^m)=(y^1(F({\phi }^{-1}(x))),\dots ,y^n(F({\phi }^{-1}(x))))\mathrm{.}$

• The domain is written explicitly. The restriction to $\phi (U\cap F^{-1}(V))$ is exactly where the composite formula makes sense; on the remaining part of $\phi (U)$ the image of ${\phi }^{-1}$ need not even lie in $V$. This restriction is the precise content of "the representative with respect to these two charts".

• Continuity is what makes the domain open. Without continuity of $F$ the set $U\cap F^{-1}(V)$ can fail to be open, and the representative is not a map between open subsets of Euclidean spaces; the definition of smoothness below therefore applies the representative only to continuous maps.

Let $m\ge 1$, let $q\in {\mathbb{N}}_0$, let $W\subseteq {\mathbb{R}}^m$ be open and let $f:W\to {\mathbb{R}}^q$. The map $f$ is smooth (of class $C^{\mathrm{\infty }}$) when every component $f_l:W\to \mathbb{R}$, $l<q$, is of class $C^k$ for every $k\in \mathbb{N}$, where scalar $C^k$ means that every iterated coordinate derivative through order $k$ exists and is continuous on $W$ ($C^k$ maps and multi-index derivative notation in Euclidean space). For $m=0$ the domain is the one-point space or the empty set and every map from it is declared smooth; for $q=0$ there are no components and every map into the one-point space is smooth.

Let $M$ be a topological $n$-manifold and let $(U,\phi )$ and $(V,\psi )$ be charts on $M$ (Manifold charts, coordinate domains, and coordinate functions). The charts are smoothly compatible when $U\cap V=\mathrm{\varnothing }$, or when $n\ge 1$ and both transition maps

$\psi \circ {\phi }^{-1}:\phi (U\cap V)⟶\psi (U\cap V),\phi \circ {\psi }^{-1}:\psi (U\cap V)⟶\phi (U\cap V)$

are smooth in the sense above. Both directions of the transition are part of the definition; in dimension zero overlapping charts have the same one-point image, the only transition is the identity, and overlapping charts are declared compatible.

• Smoothness is a property of pairs of charts, not of a chart alone. One chart has no smoothness condition: any homeomorphism onto an open set is a chart. Smoothness enters only when two charts must agree on their overlap.

• Both transition directions are required by definition. A bijective map whose one direction is smooth need not have a smooth inverse (for example, the two charts $(\mathbb{R},\mathrm{id})$ and $(\mathbb{R},x\mapsto x^3)$ have reverse transition $x\mapsto x^{1\mathrm{/}3}$, which is not differentiable at $0$); requiring both directions outright makes compatibility genuinely symmetric, as Smooth chart compatibility is symmetric and reflexive records.

Let $M$ be a topological manifold and let $(U,\phi )$ and $(V,\psi )$ be charts on $M$. Then:

1. Every chart is smoothly compatible with itself.
2. If $(U,\phi )$ and $(V,\psi )$ are smoothly compatible, then $(V,\psi )$ and $(U,\phi )$ are smoothly compatible.

Let $M$ be a topological $n$-manifold. A smooth atlas on $M$ is a family $\mathcal{A}=(U_i,{\phi }_i{)}_{i\in I}$ of charts on $M$ (Manifold charts, coordinate domains, and coordinate functions) such that:

1. the coordinate domains cover $M$, that is ${\bigcup }_{i\in I}{U}_i=M$; and
2. any two members of $\mathcal{A}$ are smoothly compatible (Smoothly compatible charts and the smoothness of Euclidean transition maps).

Two atlases $\mathcal{A}$ and $\mathcal{B}$ on $M$ are compatible when every chart of $\mathcal{A}$ is smoothly compatible with every chart of $\mathcal{B}$. The family of all charts of both atlases is written $\mathcal{A}\cup \mathcal{B}$.

• The covering condition is part of being an atlas. A family of pairwise compatible charts that does not cover $M$ is not an atlas, and the maximal atlas of Each smooth atlas is contained in a unique maximal smooth atlas still covers $M$ for exactly this reason.

• Compatibility of atlases is cross-pairwise, not pairwise within the union. Compatibility within each atlas is already given; the extra content of "compatible atlases" is the smoothness of every transition across the two families, which is the hypothesis The union of two compatible smooth atlases is a smooth atlas turns into an atlas again.

Let $\mathcal{A}$ and $\mathcal{B}$ be compatible smooth atlases on a topological manifold $M$. Then the family $\mathcal{A}\cup \mathcal{B}$ of all charts belonging to $\mathcal{A}$ or to $\mathcal{B}$ is again a smooth atlas on $M$.

Let $M$ be a topological manifold.

1. Compatibility of smooth atlases on $M$ is an equivalence relation.
2. Let $W\subseteq {\mathbb{R}}^a$, $W^{\mathrm{\prime }}\subseteq {\mathbb{R}}^b$, $W^{\mathrm{\prime }\mathrm{\prime }}\subseteq {\mathbb{R}}^c$ be open. If $u:W\to W^{\mathrm{\prime }}$ is smooth and $g:W^{\mathrm{\prime }}\to W^{\mathrm{\prime }\mathrm{\prime }}$ is of class $C^r$ ($r\in \mathbb{N}\cup \{\mathrm{\infty }\}$), then $g\circ u$ is $C^r$; and if $h:W\to W^{\mathrm{\prime }}$ is of class $C^r$ and $v:W^{\mathrm{\prime }}\to W^{\mathrm{\prime }\mathrm{\prime }}$ is smooth, then $v\circ h$ is $C^r$.

Let $\mathcal{A}$ be a smooth atlas on a topological manifold $M$. Then the set ${\mathcal{A}}^{\mathrm{\prime }}$ of all charts on $M$ that are smoothly compatible with every chart of $\mathcal{A}$ is again a smooth atlas on $M$. It contains $\mathcal{A}$, and every member of ${\mathcal{A}}^{\mathrm{\prime }}$ is by construction compatible with every chart of $\mathcal{A}$.

Let $\mathcal{A}$ be a smooth atlas on a topological manifold $M$ (Smooth atlases). The smooth structure generated by $\mathcal{A}$ is the set

$[\mathcal{A}]:=\{(U,\phi ):(U,\phi )\text{ is a chart on }M\text{ and }(U,\phi )\text{ is smoothly compatible with every chart of }\mathcal{A}\}\mathrm{.}$

By All charts compatible with a smooth atlas form a smooth atlas, $[\mathcal{A}]$ is again a smooth atlas on $M$, and since every chart of $\mathcal{A}$ is compatible with itself and with the other charts of $\mathcal{A}$, the inclusion $\mathcal{A}\subseteq [\mathcal{A}]$ holds. A chart is said to be compatible with the atlas $\mathcal{A}$ exactly when it belongs to $[\mathcal{A}]$. In the notation $[\mathcal{A}]$ the brackets name the whole generated structure, not a quotient; distinct atlases may generate the same structure, and Each smooth atlas is contained in a unique maximal smooth atlas records exactly when that happens.

• The construction is formulaic, not a choice. $[\mathcal{A}]$ is the single family of all charts satisfying the stated compatibility condition; no representative is selected from it, and no choice principle is used.

• The bracket is the maximal atlas. By its definition $[\mathcal{A}]$ already contains every chart compatible with $\mathcal{A}$, so nothing can be added to it without losing pairwise compatibility; the maximal-atlas theorem states this precisely.

Let $\mathcal{A}$ be a smooth atlas on a topological manifold $M$, with generated smooth structure $[\mathcal{A}]$.

1. $[\mathcal{A}]$ is a smooth atlas containing $\mathcal{A}$, and it is maximal: every smooth atlas on $M$ that contains $[\mathcal{A}]$ equals $[\mathcal{A}]$.
2. For a second smooth atlas $\mathcal{B}$ on $M$, the two generated structures coincide, $[\mathcal{A}]=[\mathcal{B}]$, if and only if $\mathcal{A}\cup \mathcal{B}$ is a smooth atlas.
3. Consequently $\mathcal{A}$ is contained in exactly one maximal smooth atlas, namely $[\mathcal{A}]$.

A smooth $n$-manifold is a pair $(M,\mathcal{S})$ in which $M$ is a topological $n$-manifold (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces) and $\mathcal{S}$ is a smooth structure on $M$: a maximal smooth atlas (Each smooth atlas is contained in a unique maximal smooth atlas). Because Each smooth atlas is contained in a unique maximal smooth atlas sends every smooth atlas to the unique maximal atlas containing it, a smooth manifold is equivalently specified by a topological manifold $M$ together with any one smooth atlas $\mathcal{A}$, the structure being the generated $[\mathcal{A}]$. A chart $(U,\phi )\in \mathcal{S}$ is called a smooth chart (or a chart of the smooth structure); its domain is a coordinate domain and its coordinate functions are smooth coordinates on $U$. When the structure is clear from context, the manifold itself is written $M$ in place of $(M,\mathcal{S})$.

• The maximal atlas is the structure. Two smooth atlases present the same smooth manifold exactly when they generate the same maximal atlas, which by Each smooth atlas is contained in a unique maximal smooth atlas holds exactly when their union is again a smooth atlas.

• A smooth chart is a chart of the structure, nothing more. Membership in $\mathcal{S}$ is what licenses calling a coordinate map smooth; a chart that is merely a homeomorphism onto an open set need not be smooth relative to $\mathcal{S}$.

Let $(M,\mathcal{S})$ be a smooth $n$-manifold and let $U\subseteq M$ be open, carrying the subspace topology. Then $U$ is a topological $n$-manifold. For every smooth atlas $\mathcal{A}$ with $[\mathcal{A}]=\mathcal{S}$, the family of restricted charts

$\mathcal{A}{\mathrm{\mid }}_U:=\left\{\right(V\cap U,\phi {\mathrm{\mid }}_{V\cap U}):(V,\phi )\in \mathcal{A}\}\mathrm{.}$

is a smooth atlas on $U$, and the maximal atlas it generates is independent of the presenting atlas $\mathcal{A}$: it depends only on the structure $\mathcal{S}$. This maximal atlas is the restricted smooth structure of $U$.

Let $n\ge 1$ and let $W\subseteq {\mathbb{R}}^n$ be open. Then $W$ is a smooth $n$-manifold: the one-chart atlas $\{(W,{\mathrm{i}\mathrm{d}}_W)\}$ is a smooth atlas, and the smooth structure it generates is the one induced on the open subset $W$ of ${\mathbb{R}}^n$. A chart $(U,\phi )$ on $W$ belongs to this structure exactly when both transition maps between $\phi$ and the identity are smooth, that is, exactly when $\phi :U\to \phi (U)$ and its inverse are smooth as maps between Euclidean open sets.

Let $(M,\mathcal{S})$ and $(N,\mathcal{T})$ be smooth manifolds of dimensions $m$ and $n$. Then $M\times N$ with the product topology is a topological $(m+n)$-manifold. If $\mathcal{A}$ and $\mathcal{B}$ are smooth atlases with $[\mathcal{A}]=\mathcal{S}$ and $[\mathcal{B}]=\mathcal{T}$, then the set of product charts

$\mathcal{A}\times \mathcal{B}:=\left\{\right(V\times W,\phi \times \psi ):(V,\phi )\in \mathcal{A},(W,\psi )\in \mathcal{B}\}$

is a smooth atlas on $M\times N$, and the maximal atlas it generates is independent of the presenting atlases: it depends only on $\mathcal{S}$ and $\mathcal{T}$. This maximal atlas is the product smooth structure of $M\times N$.

Let $I$ be an at most countable set with a fixed enumeration, and for each $i\in I$ let $M_i$ be a smooth $n$-manifold presented with a fixed smooth atlas and a fixed finite or countable listing of a basis of its topology. Then the disjoint union $X:={⨆}_{i\in I}{M}_i$ with the disjoint union topology is a smooth $n$-manifold: the charts of the members transported by the canonical injections form a smooth atlas whose generated maximal atlas depends only on the smooth structures of the $M_i$. If instead only the existence of second-countable topologies on the members is assumed, then selecting one basis per member uses ${\mathrm{A}\mathrm{C}}_{\omega }$; the countability of $I$ is essential, and no claim is made for an uncountable index set.

Let $M$ and $N$ be smooth manifolds, let $F:M\to N$ be a map, let $p\in M$ and let $r\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$, where ${\mathbb{N}}_0=\{0,1,2,\dots \}$. For $m,q\in {\mathbb{N}}_0$ and an open set $E\subseteq {\mathbb{R}}^m$, a map $g:E\to {\mathbb{R}}^q$ is of class $C^r$ as follows. When $m=0$, every such map is declared $C^r$. When $m\ge 1$, each component $g_l$ with $l<q$ must be $C^r$ in the sense of $C^k$ maps and multi-index derivative notation in Euclidean space; when $q=0$ this component condition is vacuous. Suppose $F$ is continuous at $p$. Then $F$ is of class $C^r$ at $p$ when there are smooth charts $(U,\phi )$ of $M$ at $p$ and $(V,\psi )$ of $N$ at $F(p)$ with $F(U)\subseteq V$ such that the coordinate representative

$\psi \circ F\circ {\phi }^{-1}:\phi (U)⟶\psi (V)$

(The coordinate representation of a map between manifolds) is of class $C^r$ on a neighbourhood of $\phi (p)$ in this componentwise Euclidean sense. By Chart independence of $C^r$ smoothness ↗, this condition is independent of the chosen charts: if one such representative is $C^r$, then every one is, so "some charts" may be read as "any charts". A map is $C^r$ on an open set $W\subseteq M$ when it is continuous on $W$ and $C^r$ at every point of $W$. A map that is $C^r$ for every finite $r$ — equivalently $C^{\mathrm{\infty }}$ — is called smooth; the term $C^r$ map between smooth manifolds is reserved for the case where $F$ is continuous and the representative condition holds at every point of $M$.

• Continuity is part of the hypothesis, not a consequence, at this point. The representative is only a map between open Euclidean sets when $F$ is continuous, as The coordinate representation of a map between manifolds records; that Smooth maps are continuous later derives continuity from the $C^1$ representative condition does not change the definition.

• The choice of charts is discharged. The well-definedness obligation — that testing one chart pair agrees with testing every chart pair — is discharged by Chart independence of $C^r$ smoothness ↗, which is why that lemma is named in justified_by rather than deps.

Let $M$ and $N$ be smooth manifolds, let $F:M\to N$ be continuous at $p\in M$, and let $r\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$. Let $(U,\phi )$, $(U^{\mathrm{\prime }},{\phi }^{\mathrm{\prime }})$ be smooth charts of $M$ at $p$ and $(V,\psi )$, $(V^{\mathrm{\prime }},{\psi }^{\mathrm{\prime }})$ smooth charts of $N$ at $F(p)$, with $F(U)\subseteq V$ and $F(U^{\mathrm{\prime }})\subseteq V^{\mathrm{\prime }}$. If the representative $\psi \circ F\circ {\phi }^{-1}$ is of class $C^r$ on a neighbourhood of $\phi (p)$, then the representative ${\psi }^{\mathrm{\prime }}\circ F\circ {\phi }^{\mathrm{\prime }-1}$ is of class $C^r$ on a neighbourhood of ${\phi }^{\mathrm{\prime }}(p)$. Testing one chart pair therefore agrees with testing any other.