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

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.

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.
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.