What structure do smooth maps and diffeomorphisms preserve?

Question

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.

2. 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 $M$.

Answer 1

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.

Answer 2

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.