I am following the Nigel's differentiable manifolds lectures. In page 14 you can find the following definition:
A map $F:M\to N$ of manifolds is a smooth map if for each point $x\in M$ and chart $(U_\alpha,\varphi_\alpha)$ in $M$ with $x\in U$ and chart $(V_i,\psi_i)$ in $N$ with $F(x)\in V_i$, the set $F^{-1}(V_i)$ is open and the composite function
$$\psi_i\circ F\circ \varphi^{-1}_\alpha$$
on $\varphi_\alpha(F^{-1}(V_i)\cap U_\alpha)$ is a $C^\infty$ function
My question is: Why he imposes that $F^{-1}(V)$ must be open? Many other autors only requires that $\psi_i\circ F\circ \varphi^{-1}_\alpha$ must be $C^\infty$ in $\varphi_\alpha(U)$
Best Answer
I think the point is that, as Hitchin explains on p9, he is not assuming in the definition of a manifold that it is a topological space, unlike most authors. Instead he derives the topology from the definition. Hence on p14 he does not assume in Definition 6 that F is continuous. As a result he has to say explicitly that $F^{-1}(V_i)$ must be open. Other authors would assume F is continuous so it follows automatically that $F^{-1}(V_i)$ is open, as $V_i$ is open.