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 $F(U)\subseteq V$?

$\endgroup$
  • $\begingroup$ Note that $U \cap F^{-1}(V) \neq \emptyset$ always, since it contains $p$ $\endgroup$ – 57Jimmy May 15 '18 at 14:02
  • $\begingroup$ @57Jimmy: Didn't I mention it too? $\endgroup$ – user 170039 May 15 '18 at 14:02
  • $\begingroup$ Sorry, I thought you meant something else :) $\endgroup$ – 57Jimmy May 15 '18 at 14:03
  • $\begingroup$ @57Jimmy: No problem. $\endgroup$ – user 170039 May 15 '18 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 '18 at 14:04
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.

$\endgroup$
  • $\begingroup$ So you say that instead of $(U,\varphi)$ we consider $(U',\varphi|_{U'})$, right? $\endgroup$ – user 170039 May 15 '18 at 14:05
  • $\begingroup$ Exactly, that's what I meant $\endgroup$ – 57Jimmy May 15 '18 at 14:06
  • $\begingroup$ Nice. Sorry if this was a trivial question. $\endgroup$ – user 170039 May 15 '18 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 '18 at 14:08

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.