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.
Best 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.