# smooth manifolds, equivalent statements

Let $X,Y$ be smooth manifolds. Show: A function $f:X\to Y$ is smooth, iff for every open $V\subseteq Y$ and every smooth function $g:V\to\mathbb{R}$ the composition $g\circ f: f^{-1}(V)\to\mathbb{R}$ is smooth.

Hello,

I want to prove this statement. I show "$\Rightarrow$" first.

"$\Rightarrow$":

Let $f:X\to Y$ be smooth. Let $V\subseteq Y$ be open and $g:V\to\mathbb{R}$ smooth.

Observe $g\circ f: f^{-1}(V)\to\mathbb{R}$. Show: $g\circ f$ is smooth.

$g\circ f$ is smooth, when the function is smooth around every $v\in f^{-1}(V)$. Let $v\in f^{-1}(V)$ be random. Since $f$ is smooth, there is a smooth map $(U,\varphi)$ around $v$ and $(W,\psi)$ around $f(v)$ such that, $\psi\circ f\circ\varphi^{-1}$ is smooth on a neighbourhood of $\varphi(v)$.

Since $g$ is smooth there is additionally a smooth map $(W',\psi')$ around $g(f(v))$ such that, $\psi'\circ g\circ f\circ\varphi^{-1}$ is smooth on a neighbouhood of $\psi'(v)$.

Then is $g\circ f$ smooth around every $v\in f^{-1}(V)$.

As it stands, you argument does not look correct. For several statements like "there is a smooth map $(U,\phi)$" you should use local charts instead of smooth maps. Moreover, for $g$ you only need a chart in the domain space whereas you are trying to use a local chart on the target space. The main issue however is that the implication you want to prove follows from the facts that a restriction of a smooth map to an open subset is smooth and that a composition of smooth maps is smooth, which probably should be proved first.
• We know, that the composition of smooth manifolds is smooth and we also know, that when $f:X\to Y$ is smooth around $x\in X$, so is $\psi\circ f\circ \phi^{-1}$ smooth on a neighbourhood of $\phi(x)$ for every smooth maps $(U,\phi)$ around $x$ and $(V,\psi)$ around $f(x)$. – MrTopology Jun 30 '16 at 16:04
• Yes, but to conclude smoothness of $f$ from smoothness of $\psi\circ f\circ \phi^{-1}$ you need charts. – Andreas Cap Jun 30 '16 at 17:13