# If for each $\alpha, \beta$, the map $\psi_{\beta} \circ F \circ \phi_{\alpha}^{-1}$ is smooth, then $F$ is smooth.

Let $F:M\to N$ be a map. Suppose $\mathcal{A} = \{(U_{\alpha},\phi_{\alpha})\}$ and $\mathcal{B} = \{(V_{\beta},\psi_{\beta})\}$ are smooth atlas for $M$ and $N$ respectively. Suppose that for each $\alpha, \beta$, the map $\psi_{\beta} \circ F \circ \phi_{\alpha}^{-1}$ is smooth, then $F$ is smooth.

Proof:

In order to show that $F$ is smooth, we need to show that for any chart $(U,\phi)$ of $M$ and $(V,\psi)$ of $N$, the map $$\psi \circ F \circ \phi^{-1}: \phi(U\cap F^{-1}(V)) \to \psi(F(U)\cap V),$$ is smooth. Let $x = \phi(p) \in \phi(U\cap F^{-1}(V))$. Since $\mathcal{A}$ is an atlas, there exist a chart $(U_{\alpha},\phi_{\alpha})$ for which $p \in U_{\alpha}$. Also, $p \in U \cap F^{-1}(V)$ which implies that $p \in U_{\alpha} \cap F^{-1}(V)$ and thus $F(p) \in F(U_{\alpha})\cap V$. Now, $\mathcal{B}$ is an atlas for $N$, so there exists a chart $(V_{\beta},\psi_{\beta})$ for which $F(p) \in V_{\beta}$, so that $p \in F^{-1}(V_{\beta})$. It follows that $p \in U_{\alpha} \cap F^{-1}(V_{\beta})$. Now, the map $$\psi_{\beta} \circ F \circ \phi_{\alpha}^{-1} \text{ is smooth.}$$ Also, the charts $(U_{\alpha},\phi_{\alpha})$ and $(U,\phi)$ are compatible, since $\mathcal{A}$ is an smooth atlas. Similarly, the charts $(V_{\beta},\psi_{\beta})$ and $(V,\psi)$ are also compatible. It follows that both maps $$\phi_{\alpha} \circ \phi^{-1} \text{ and } \psi \circ \psi^{-1}_{\beta} \text{ are smooth}.$$ Hence, the map $$\psi \circ F \circ \phi^{-1}=\underbrace{(\psi\circ\psi_{\beta}^{-1})}_{\text{smooth}}\circ\underbrace{(\psi_{\beta} \circ F \circ \phi_{\alpha}^{-1})}_{\text{smooth}}\circ\underbrace{(\phi_{\alpha}\circ\phi^{-1})}_{\text{smooth}},$$ is smooth.

Is this proof correct? I have been reading about manifolds for about 4 days and not sure if this is the right approach.