How to see that "two manifolds are diffeomorphic when you can give them each a coordinate atlas with the same transition maps"

Question

This question is about the diffeomorphism of $\mathbb{C}P^1$ and $S^2$. At the end of youler's answer, we read

"the general fact that two manifolds are diffeomorphic when you can give them each a coordinate atlas with the same transition maps."

I am not sure why this is true. I have not even a geometric intuition.

In my lecture notes, we were given the same example, up to showing that the two manifolds have the same transition maps. I think we are expected to understand that the transition maps imply that the two manifolds must be diffeomorphic to each other, but I don't see why.

If possible, I would like an intuitive reasoning, followed by a rigorous answer.

Answer 1

Suppose that $X, Y$ are manifolds with atlases $\{\phi_\alpha: \alpha\in A\}$ and $\{\psi_\alpha: \alpha\in A\}$ such that for any two pairs of indices $\alpha, \beta\in A$, $$ \phi_\beta\circ \phi_\alpha^{-1}= \psi_\beta\circ \psi_\alpha^{-1}. $$
Define the map $f: X\to Y$ by the formula $$ f(x)=y $$ whenever $\phi_\alpha(x)= \psi_\alpha(y)$ for any pair of charts with common index whose domains contains $x$ and $y$ respectively. The fact that the transition maps of the two atlases are the same implies that this map is well-defined (i.e. are independent of the choice of the above charts). To see that this map is a diffeomorphism, one needs to check smoothness of this map (and of its inverse) in local coordinates, i.e. smoothness of compositions $$ \psi_\alpha \circ f \circ \phi_\alpha^{-1} $$ and $$ \phi_\alpha \circ f^{-1} \circ \psi_\alpha^{-1}. $$ But these compositions are equal the identity maps for every $\alpha\in A$. Hence, $f$ and its inverse are diffeomorphisms. Thus, $f: X\to Y$ is a diffeomorphism.

The converse statement holds as well: Suppose that there is a diffeomorphism $f: X\to Y$. Then $X, Y$ can be equipped with atlases that have the same transition maps. Indeed, given an atlas $\{\psi_\alpha: \alpha\in A\}$ on $Y$, define charts on $X$ by the formula $$ \phi_\alpha= \psi_\alpha\circ f. $$ Direct computations shows that transition maps for the two atlases are the same.

Answer 2

Consider two manifolds $M$ and $N$, with atlases $\{(U_{\alpha}, \phi_{\alpha})\}$ and $\{(V_{\alpha}, \psi_{\alpha})\}$ respectively. Let $\tau = \phi_{\beta} \circ\phi_{\alpha}^{-1}:\phi_{\alpha}(U_{\alpha}\cap U_{\beta})\rightarrow\phi_{\beta}(U_{\alpha}\cap U_{\beta})$ and $\xi=\psi_{\beta} \circ\psi_{\alpha}^{-1}:\psi_{\alpha}(V_1\cap V_{\beta})\rightarrow\psi_{\beta}(V_{\alpha}\cap V_{\beta})$ be the transition maps associated with each atlas.

If $\tau = \phi_{\beta} \circ\phi_{\alpha}^{-1} = \psi_{\beta} \circ\psi_{\alpha}^{-1} = \xi$, we then have $$\phi_{\beta}(U_{\alpha}\cap U_{\beta})=\psi_{\beta}(V_{\alpha}\cap V_{\beta}) \implies M \supset U_{\alpha}\cap U_{\beta}=V_{\alpha}\cap V_{\beta} \subset N,$$ since both $\phi_{\beta}$ and $\psi_{\beta}$ are homeomorphisms.

Let's denote this open set with $S=U_{\alpha}\cap U_{\beta}=V_{\alpha}\cap V_{\beta}$.

As stated in youler's answer, consider the map $\Phi:M \rightarrow N$, defined by $\Phi(\phi_{\alpha}^{-1}(S))=\psi_{\alpha}^{-1}(S)$. Its inverse $\Phi^{-1}:N \rightarrow M$, also exists and is defined by $\Phi(\psi_{\alpha}^{-1}(S))=\phi_{\alpha}^{-1}(S)$. If we can somehow show that $\Phi$ and $\Phi^{-1}$ are both smooth... then indeed the two manifolds are diffeomorphic.