2
$\begingroup$

This is Exercise 3.1.A. in Vakil's notes

Suppose that $\pi: X\rightarrow Y$ is a continuous map of differentiable manifolds. Show that $\pi$ is differentiable if differentialble functions pull back to differentiable functions, i.e., if pullback by $\pi$ gives mpa $\mathcal{O}_Y\rightarrow \pi_*\mathcal{O}_X$.

Let $f: V\subseteq Y\rightarrow \mathbb{R}$ be a differentiable function on an open subset of $Y$. Let $(U,\phi)$ be a chart of $X$, $(V,\psi)$ be a chart of $Y$. Then $f\circ\pi(\phi^{-1})$ is differentiable and $f\circ\psi^{-1}$ is differentiable. We need to show that $\psi\circ\pi\circ\phi^{-1}$ is differentiable. I don't know how to prove this.

Can we say that since $(U,\psi\circ\pi)$ constructs a chart of $X$, by the compatibility, $\psi\circ\pi\circ\phi^{-1}$ is differentiable? This seems true but this does not use the fact that $f$ is differentiable.

Sorry I have no background of Differentiable Manifolds. Any help would be appreciated.

I also saw this post: pullback of continuous maps of manifolds, but I don't understand how $\gamma$ is smooth in the answer. $\gamma=\beta\circ\rho$, where $\rho$ is defined to be smooth, but $\beta$ is not.

$\endgroup$
  • $\begingroup$ I don't change the names: the condition "$\psi\circ\pi\circ\phi^{-1}$ is differentiable for any charts $(U,\phi)$ and $(V,\psi)$" is the definition of differentiable map $\pi:X\to Y$! $\endgroup$ – Armando j18eos Mar 19 '17 at 11:47
  • $\begingroup$ @Armandoj18eos: But $\pi$ is not differentiable yet. Isn't that what we need to show? Also $\psi$, $\phi$ are not differentiable. They are just homeomorphisms, right? $\endgroup$ – KittyL Mar 19 '17 at 12:02
  • $\begingroup$ @Armandoj18eos: Thanks! I see! $\psi$ is differentiable since differentiable functions are defined as a function whose compositions with inverse of all charts are differentiable. I didn't realized this before. $\endgroup$ – KittyL Mar 19 '17 at 12:50
1
$\begingroup$

Let $\pi:X\to Y$ be a continuous map of differentiable manifolds such that \begin{equation*} \forall V\subseteq Y\,\text{open,}\,\pi_{*}(V):f\in\mathcal{O}_Y(V)\mapsto f\circ\pi_{|\pi^{-1}(V)}\in\mathcal{O}_X(\pi^{-1}(V)); \end{equation*} if $(V,\psi)$ is a chart of $Y$, without loss of generality one can assume that $(\pi^{-1}(V)=U,\phi)$ is a chart of $X$.

By definition $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$, then $\psi\circ\pi_{|U}\circ\phi^{-1}:\mathbb{R}^n\to\mathbb{R}^m$ is a continuous map; by hypothesis \begin{equation*} \forall f\in\mathcal{O}_Y(V),\,f\circ\pi_{|U}=(f\circ\psi^{-1})\circ(\psi\circ\pi_{|U}):U\to V\to\mathbb{R} \end{equation*} is differentiable, that is \begin{equation*} f\circ\pi_{|U}\circ\phi^{-1}=(f\circ\psi^{-1})\circ(\psi\circ\pi_{|U}\circ\phi^{-1}):\mathbb{R}^n\to U\to\mathbb{R} \end{equation*} is differentiable.

Passing to coordinates, for any $f\in\mathcal{O}_X(V)$ is \begin{equation*} f\circ\psi^{-1}:\underline{y}=(y_1,\dots,y_m)\in\mathbb{R}^m\to f\left(\psi^{-1}\left(\underline{y}\right)\right)\in\mathbb{R}, \end{equation*} in other words, $f\circ\psi^{-1}$ is a differentiable function in $m$ variables. Of course, the $y_i$'s are the (local) coordinates of $Y$ on $V$; one kows that the functions \begin{equation*} \psi_i\equiv pr_i\circ\psi^{-1}:(y_1,\dots,y_m)\in\mathbb{R}^m\to y_i\in\mathbb{R} \end{equation*} are differentiable.

The same reasoning holds for $\phi$!

By previous remark, one proves that the component functions of $\pi_{|U}$ are differentiable; by gluing axiom, $\pi$ is a differentiable map of manifolds.

$\endgroup$
  • $\begingroup$ In fact, we don't need $f$ here, right? If the component functions of $\psi$ are differentiable, then $\psi\circ\pi\circ\phi^{-1}$ is differentiable, which proves $\pi$ is differentiable. $\endgroup$ – KittyL Mar 19 '17 at 17:07
  • $\begingroup$ No, we need of $f$; because the differentiability of $\psi_i$'s (and $\phi_j$'s) is proved using the differentiable curve on $Y$ (and $X$). Do you understand me? $\endgroup$ – Armando j18eos Mar 20 '17 at 11:09
  • $\begingroup$ Sorry I am not familiar with this subject. But I found that the definition of a differentiable function, say $\alpha$, on a differentiable manifold $M$ is: for any chart $(U,\phi)$ on $M$, $\alpha\circ\phi^{-1}$ is differentiable. But doesn't this imply that all charts must be differentiable since they satisfy the compatibility property? So in our case, shouldn't $\psi_j$ be automatically differentiable? Thank you for your answer. $\endgroup$ – KittyL Mar 20 '17 at 12:21
  • $\begingroup$ Sorry for my absence; just a remark: $\psi$ need not to be differentiable, see here! $\endgroup$ – Armando j18eos Mar 22 '17 at 17:29
  • 1
    $\begingroup$ No, it's an buse of notation, as I edited! ;) $\endgroup$ – Armando j18eos Mar 24 '17 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.