I just want to double check that I established that diffeomorphism correctly.
First, let's recover the definition of a diffeomorphism $F$ between two smooth manifolds $M$ and $N$. We say that $F:M\to N$ is a diffeomorphism if $F$ is bijective, $F$ is smooth and $F^{-1}$ is smooth.
Consider the map $F:S^2\to\mathbb{CP}^1$ where
$$F(x,y,z)=\begin{cases}
[1;\frac{x}{1-z}+\frac{y}{1-z}i], & \text{if}\ (x,y,z)\neq (0,0,1) \\
[0;1], & \text{otherwise}
\end{cases}$$
and the map $F^{-1}:\mathbb{CP}^1\to S^2$ where
$$F^{-1}([1,u+vi])=(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1})$$
and $F^{-1}([0;1])=(0,0,1)$. We can easily see that $F$ is bijective as $F\circ F^{-1}=F^{-1}\circ F=id$.
Next, consider the stereographic projection charts $\{U_0,U_1\}$ for $S^2$ and the standard charts $\{V_0,V_1\}$ for $\mathbb{CP}^1$ where, for example, $V_0=\{[z_0,z_1]|z_0\neq 0\}$ with the corresponding homeomorphisms $\phi_i$ and $\psi_i$, $i=0,1$. To show that $F$ is smooth we want to show that
$$\psi_i\circ F\circ \phi^{-1}_j:\phi_j(U_j\cap F^{-1}(V_i))\to V_i\text{ is a smooth map, for all }i,j$$
which follows from the basic computations. For example,
$$\psi_1\circ F\circ \phi_0^{-1}(u,v)=(\frac{u}{u^2+v^2},\frac{-v}{u^2+v^2}).$$
Is there another faster way to establish a diffeomorphism?
Best Answer
The proof looks fine, except that you skipped proving smoothness of $F^{-1}$.
Regarding your final question, let me ask my own (rhetorical) question: What part of the proof would you want to skip?
Generally speaking, to prove two manifolds are diffeomorphic requires you to write down a formula of a diffeomorphism $F$ and its inverse map $F^{-1}$, and then to prove all the required properties (smoothness; compositions are the identity). Which is just what you have written out, except for smoothness of $F^{-1}$.
Now, sometimes there are shortcuts that you can take to proving two manifolds $X,Y$ are diffeomorphic, which do not require you to write down any new formulas.
For example, perhaps you can track down a sequence of theorems regarding diffeomorphisms $X=X_0 \cong X_1 \cong \cdots \cong X_k=Y$; you may now conclude that $X$ and $Y$ are diffeomorphic simply by citing those theorems.
Or, perhaps there is a big classification theorem, for example that compact, connected, oriented 2-manifolds are completely classified up to diffeomorphism by their genus. Perhaps, given two compact, connected, oriented 2-manifolds $X,Y$, you can compute the genus of each and check that those genuses are equal, and now you may conclude that $X$ and $Y$ are diffeomorphic by citing the classification theorem.
Even in those cases, though, somewhere at the root of the previous diffeomorphism theorems you will find actual constructions of actual diffeomorphisms.