I'm reading up about pullbacks and pushforwards. If $\psi:M\to M'$ is smooth and $f:M'\to\mathbb{R}$ is a smooth function, then we define the pullback as $\psi*f=f\circ\psi$. By definition of a smooth map between manifolds then, the pullback of $f$ is also a smooth function on $M$.
Even though this is clear, I'm still trying to understand this in terms of charts. Let $M,M'$ be $n,n'$ dimensional manifolds respectively. In terms of charts, the definition of a smooth map $\psi:M\to M'$ is one for which $\phi'\circ\psi\circ\phi^{-1}:\mathbb{R}^n\to\mathbb{R}^{n'}$ is smooth, for homeomorphisms $\phi,\phi'$ corresponding to arbitrary charts in $M,M'$ respectively.
Also, a smooth function $f:M'\to\mathbb{R}$ can be defined as one for which $f\circ\phi'^{-1}:\mathbb{R}^{n'}\to\mathbb{R}$ is smooth for all homeomorphisms corresponding to charts covering $M'$.
With this, I can rephrase the conditions in bold at the top. If I have a map $\psi:M\to M'$ such that $\phi'\circ\psi\circ\phi^{-1}$ is smooth for all $\phi,\phi'$ and if there exists some $f:M'\to\mathbb{R}$ such that $f\circ\phi'^{-1}$ is smooth for all $\phi'$, then to show that the pullback is also smooth (in terms of charts), I have to show that $(f\circ\psi)\circ\phi^{-1}$ is smooth for all chart homeomorphisms $\phi$ on $M$.
Is this alternative statement of the problem correct? And how do I prove this? i.e. how to show that smoothness of $\phi'\circ\psi\circ\phi^{-1}$ and smoothness of $f\circ\phi'^{-1}$ for all $\phi,\phi'$ implies smoothness of $f\circ\psi\circ\phi^{-1}$ for all $\phi$?
Best Answer
Note that, restricted to the appropriate domain, we have $f\circ\psi\circ\phi^{-1} = (f\circ\phi'^{-1})\circ(\phi'\circ\psi\circ\phi^{-1})$ for any $\phi'$. As the two paranthetical terms are smooth, so is their composition. Smoothness is a local condition, and the above holds for every $\phi'$, so it follows that $f\circ\psi\circ\phi^{-1}$ is smooth, and hence $\psi^*f = f\circ\phi$ is a smooth function on $M$.