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Context:

  1. Continuous maps between topological spaces are structure preserving in the following sense:

Given two topological spaces $(X,\tau_X),(Y,\tau_Y)$ (where $(X,\tau_X)$ is the topological space of set $X$ endowed with a topology $\tau_X$; similarly $(Y,\tau_Y)$), a map $f:(X,\tau_X)\to (Y,\tau_Y)$ is continuous iff $f^{-1}(V)\subset X$ is open whenever $V\subset Y$ open.

  1. A smooth manifold is a Hausdorff second countable locally Euclidean space with a smooth structure. The smooth structure is one where you take a smooth atlas and consider the unique maximal smooth atlas generated by it.

Problem:

Question 1: What is the structure on a smooth manifold? Is it this maximal smooth atlas itself or merely the requirement that the co-ordinate charts (which the underlying topological manifold already has) need to be smoothly compatible?

Question 2: I want to show or think of smooth maps as the maps that preserve this structure of smooth manifolds. How can I do that?

Here is my (WRONG) guess: given a map $f:M\to N$ ($M,N$ smooth manifolds), $f$ is smooth iff for every smooth chart $(V,\psi)$ in the smooth structure of $N$, we get a unique smooth chart $(U,\phi)$ (via some kind of taking pre-images) in the smooth structure of $M$.

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2 Answers 2

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So first things first, every diffeomorphism is also a homeomorphism by definition, so diffeomorphisms automatically preserve every topological property about manifolds.

Since Lee has already given you an excellent answer, I will provide an alternative but equivalent view. Apologies if you know nothing about sheafs.

Now a smooth manifold also comes equipped with it's sheaf of smooth functions $C^\infty_M$, that is for each open set $U\subset M$ we take continuous functions $f:U\rightarrow \mathbb R$ such that in every coordinate chart $(V\subset U,\phi)$, $f\circ \phi^{-1}$ is a smooth function on an open subset of $\mathbb R^n$ . A continuous map $F:M\rightarrow N$ is then smooth if and only if for every open set $U\subset N$, we have that: \begin{align} F^\sharp:C_N^\infty(U)&\longrightarrow C^\infty_M(F^{-1}(U))\\ f&\longmapsto f\circ F \end{align} is a morphism of rings which commutes with restriction maps. In other words, we have that $F$ is smooth if and only if this prescription above defines a sheaf morphism: $$F^\sharp:C^\infty_N\longrightarrow F_*C^\infty_M$$ This all follows pretty much directly from the definition of a smooth map, just rephrased in fancier language. However, this has the added benefit of telling us that $F$ is a diffeomorphism if and only if $F$ is a homeomorphism and $F^\sharp$ is an isomorphism of sheaves.

In fact more is true, but I believe what I am about to say is hard to prove. Two manifolds $M$ and $N$ are diffeomorphic if and only if the ring of global smooth functions, $C^\infty(M)$, is isomorphic to $C^\infty(N)$. So in a sense diffeomorphisms can be seen as the smooth maps between manifolds which induce isomorphisms on the rings of global smooth functions.

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  • $\begingroup$ This is very informative, thanks :)) I have a basic understanding of sheafs and the need for ringed spaces (from Wedhorn's Manifolds, sheaves and cohomology). It is somewhat more satisfying to see $F^\#$. $\endgroup$
    – frelg
    Feb 25 at 14:21
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To answer question (1), it is the atlas itself which defines the smooth structure. Given a smooth manifold $M$ with its atlas that I will denote $\mathcal A$, it is quite possible that $M$ has another smooth structure with atlas denoted $\mathcal A'$, but that $\mathcal A$ and $\mathcal A'$ are not compatible with each other in the sense that the overlap map between some chart in $\mathcal A$ and some chart in $\mathcal A'$ is not smooth; if that happened, we would say that $\mathcal A$ and $\mathcal A'$ define distinct smooth structures on $M$. Even on the real number line $M=\mathbb R$ this is possible: take $\mathcal A$ to have a unique chart $(U,\phi)$ where $U=\mathbb R$ and $\phi : \mathbb R \to \mathbb R$ is the identity map $\phi(x)=x$; and take $\mathcal A'$ to have also a unique chart $(V,\psi)$ where $V=\mathbb R$ and $\psi : \mathbb R \to \mathbb R$ is the cubing map $\psi(x)=x^3$. The overlap map $\phi(\psi^{-1}(y)) = \sqrt[3]{y}$ is not smooth.

The correct answer to question (2) is reasonably simple: smoothness of $f$ is defined by smoothness of any coordinate representation of $f$. That is, for any $x \in M$ with $y=f(x) \in N$, and for any chart $(U,\phi)$ of $M$ around $x$, and for any chart $(V,\psi)$ of $N$ around $y$, there is an associated coordinate representation $F$ of the map $f$ defined on $\phi(f^{-1}(V))$ by the formula $F(p) = \psi(f(\phi^{-1}(p)))$. To say that $f$ is smooth means that $F$ is smooth, for any choice of $x$ and $(U,\phi)$.

You already know that your guess is wrong, but let me emphasize an issue. One needs a definition of smoothness with several features: the definition should be applicable for maps between manifolds of different dimensions; and even if the two manifolds have the same dimension it should be applicable for functions which are not injective, or are not surjective. It's not at all clear how, in your (wrong) guess, one gets another chart $(U,\phi)$ in such situations.

Edited, in light of the comments:

If you want to define what it means for a bijection $f : M \to N$ to be a diffeomorphism, then your definition easily adapts to become quite correct. To say that $f$ is a diffeomorphism is equivalent to the statement that for every chart $(V,\psi)$ in the given atlas on $N$, the chart $(U,\phi)=(f^{-1}(V),\psi \circ f)$ on $M$ is compatible with the given atlas on $M$, and also for every chart $(U,\phi)$ in the given atlas on $M$ the chart $(V,\psi)=(f(U),\phi \circ f^{-1})$ is compatible with the given atlas on $N$. In this situation it would also be completely fair to say that for $f$ to be a diffeomorphism means that it preserves the smooth structure.

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    $\begingroup$ @frelg Perhaps the connotation you imagine for "preserving" is too strong. It might be more accurate to say "compatible" instead (esp. in light of the perspective in Chris' answer). The terminology "preserving" seems best applicable when the additional structure can always be pulled back or pushed forward along more general maps, but that's not always the case. $\endgroup$
    – Thorgott
    Feb 25 at 13:54
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    $\begingroup$ @frelg I believe pullbacks only exist when both smooth maps $M\rightarrow X$, and $N\rightarrow X$ are submersions (they might need to be surjective as well). The basic subcategory to work in is to work with is smooth fibre bundles over some base manifold $X$, then you can take fibre products/pullbacks over $X$. $\endgroup$
    – Chris
    Feb 25 at 14:09
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    $\begingroup$ After the comments here regarding "preserving" versus "compatible", I edited my answer to address these comments with regard to the concept of diffeomorphism. $\endgroup$
    – Lee Mosher
    Feb 25 at 14:36
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    $\begingroup$ Also, regarding your very first comment above, the intuition that morphisms be "structure preserving", or even that they be "structure compatible", is a rather vague intuition that works well in a lot of situations but not always so sell in others. I can only suggest that after one sees about a trillion categories, the ins and outs of the intution become manageable (and from a personal viewpoint, I keep learning new ins and new outs). $\endgroup$
    – Lee Mosher
    Feb 25 at 14:39
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    $\begingroup$ @frelg The point that was made in the Edit is pretty much exactly what I had in mind with my insufficiently worded first comment regarding when "preserving" seems more appropriate. As for pullbacks/pushouts, they indeed typically don't exist. There are at least some well-understood special cases where they do, however: the pullback of cospan in which the legs are transverse to another exists and the pushout of a span in which the legs are each open embeddings exists. $\endgroup$
    – Thorgott
    Feb 25 at 14:45

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