Let $M$ and $N$ be two smooth manifolds, let $F: M \to N$ be a map between those two manifolds. $F$ is said to be smooth (or differentiable) if $\psi \circ F \circ \phi^{-1}$ is smooth, for every $(U, \phi)$ local chart on $M$ and for every $(V, \psi)$ local chart on $N$. Show that $F$ is smooth if and only if for all $p \in M$, there exists a neighbourhood $U$ of $p$ such that $F_{|U}$ (restriction of $F$ to $U$) is smooth.
My problem with this question is that this seems to be trivially obvious to me, and I am struggling a bit to formalise the proof. The $\implies$ implication is easy, as it suffices to take $M$ as the neighbourhood $U$. As for the viceversa.. well, saying that $F_{|U}$ is smooth means that $\psi \circ F_{|U} \circ \phi^{-1}$ is smooth for every local chart. Now, since this holds for all $p \in M$, this means that the neighbourhoods $U$ form an open covering of the manifold $M$. The restriction of $F$ to each element of this covering is smooth. It looks obvious to me that this implies that $F$ is smooth, but I can't figure out a way of showing this explicitly. Any hint is appreciated.
Best Answer
The key point is that smoothness of maps of Euclidean spaces is a local property: for an open set $\Omega \subset \mathbb R^m$, we say $f : \Omega \to \mathbb R^n$ is smooth if it is smooth at all $x \in \Omega$, where this means the derivatives $D^k f(x)$ exist for all $k \in \mathbb N.$ Since (the existence of) $D^k f(x)$ is determined by the restriction of $f$ to any neighbourhood of $x,$ knowing $f|_U$ (for some open $U \ni x$) is smooth at $x$ implies that $f$ is smooth at $x$. Thus if we have such a neighbourhood $U$ for each $x \in \Omega,$ we see that $f$ is smooth at each point in $\Omega$, and thus smooth on $\Omega.$
Once we have this, transferring it to the manifold setting is an exercise in dealing with charts. Assume $f: M \to N$ is locally smooth (the RHS of your equivalence), and let $(U,\phi), (V,\psi)$ be charts as in your definition, so that we need to show $$F := \psi \circ f \circ \phi^{-1} : \phi(U) \to \psi(V)$$ is smooth. For each $x = \phi(p) \in \phi(U),$ we know there is an open neighbourhood $E$ of $p \in M$ such that $f|_E$ is smooth. The restriction $(E \cap U,\phi|_{E \cap U})$ is a chart for $E,$ so by the definition of smoothness (for maps between manifolds) we know that $\psi \circ f \circ (\phi|_{E \cap U})^{-1} : \phi(E \cap U) \to \psi(V)$ is smooth. But this function is just $F|_{E \cap U},$ so we have shown that each $x \in \phi(U)$ has a neighbourhood $E \cap U$ on which $F$ is smooth, implying $F$ is smooth.