0
$\begingroup$

Problem

Let $M$ and $N$ be smooth manifolds and let $F: M\to N$ be any map. Then prove that the followings are equivalent,

  1. For every $p \in M$ there exists smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $F(U)\subseteq V$ and the composite map $\psi\circ F\circ \varphi^{-1}$ is smooth from $\varphi(U)$ to $\psi(V)$.

  2. For each $p\in M$ there exist smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $U\cap F^{-1}(V)$ is open in $M$ and the composite map $\psi\circ F\circ \varphi^{-1}$ is smooth from $\varphi(U\cap F^{-1}(V))$ to $\psi(V)$,

My Attempt

I think that $(1)\implies (2)$ is easy. It is so because if we assume $(1)$ then for every $p\in M$, there exists smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $F(U)\subseteq V$. Now $$F(U)\subseteq V\implies F^{-1}(F(U))\subseteq F^{-1}(V)\implies U\subseteq F^{-1}(F(U))\subseteq F^{-1}(V)$$ and hence $U\cap F^{-1}(V)=U$, which is open in $M$. Thus we have shown that for every $p \in M$ there exists smooth charts $(U,\varphi)$ containing $p$ and $(V,\psi)$ containing $F(p)$ such that $U\cap F^{-1}(V)$ is open in $M$ and the composite map $\psi\circ F\circ \varphi^{-1}$ is smooth from $\varphi(U\cap F^{-1}(V))$ to $\psi(V)$. hence we are done.

Where I am stuck

To prove that $(2)\implies (1)$, I think it is enough to show the existence of a chart $(V,\psi)$ containing $F(p)$ such that $F(U)\subseteq V$ because then $U\cap F^{-1}(V)=U$. However, to prove this the problem I am facing is that the hypothesis of $(2)$ implies that $U\cap F^{-1}(V)\ne \emptyset$. But how can I use it to show that $F(U)\subseteq V$?

$\endgroup$
5
  • $\begingroup$ Note that $U \cap F^{-1}(V) \neq \emptyset$ always, since it contains $p$ $\endgroup$
    – 57Jimmy
    May 15, 2018 at 14:02
  • $\begingroup$ @57Jimmy: Didn't I mention it too? $\endgroup$
    – user170039
    May 15, 2018 at 14:02
  • $\begingroup$ Sorry, I thought you meant something else :) $\endgroup$
    – 57Jimmy
    May 15, 2018 at 14:03
  • $\begingroup$ @57Jimmy: No problem. $\endgroup$
    – user170039
    May 15, 2018 at 14:03
  • 1
    $\begingroup$ In general, though, you cannot expect $F(U) \subseteq V$ ($U$ could even be the whole space), that's why you have to change the chart. The $U$ and $V$ of $(1)$ are not necessarily the same as the $U$ and $V$ of $(2)$ $\endgroup$
    – 57Jimmy
    May 15, 2018 at 14:04

1 Answer 1

2
$\begingroup$

$(1) \Rightarrow (2)$ is correct. Hint for $(2) \Rightarrow (1)$: restrict the chart $U$ to $U':=U \cap F^{-1}(V)$. The important thing is that there is some chart, so you can use $U'$ instead of $U$, since an open subset of a chart is still a chart.

$\endgroup$
4
  • $\begingroup$ So you say that instead of $(U,\varphi)$ we consider $(U',\varphi|_{U'})$, right? $\endgroup$
    – user170039
    May 15, 2018 at 14:05
  • $\begingroup$ Exactly, that's what I meant $\endgroup$
    – 57Jimmy
    May 15, 2018 at 14:06
  • $\begingroup$ Nice. Sorry if this was a trivial question. $\endgroup$
    – user170039
    May 15, 2018 at 14:06
  • $\begingroup$ It was not: manifolds are a complicated object to study at the beginning, because of the many possible definitions of everything. $\endgroup$
    – 57Jimmy
    May 15, 2018 at 14:08

Your Answer

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