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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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