Question: Let $X^m$ and $Y^n$ differentiable manifolds. $f:X\rightarrow Y$ a differentiable map. Show that if $\partial X=\emptyset$, $y\in Reg(f)$ and $f^{-1}(y)\neq\emptyset$, then $y\not\in\partial Y.$


If there is $y\in\partial Y.$ Then if $F:U\subset\mathbb H^n\rightarrow V$, $p\in U$ such that $F(p)=y$, then $p\in \partial \mathbb H^n$.

Let $x\in f^{-1}(y)$, my idea is to proof that $x\in \partial X$, which is a contradiction.

As $f$ is differentiable, let $G:W\subset\mathbb R^m\rightarrow Z$ a parametrization of $X$, $q\in W$ with $G(q)=x$. Then, $F^{-1}\circ f\circ G$ is differentiable at $q$.

And what else can i do?


1 Answer 1


Hint: Think about what the map $f$ does to points around $x$. Also, recall what a submersion means for the dimensions of your two spaces.


Everything you did was correct. You just need to push the conclusion out of what you already know. And that comes from the submersion theorem. So we can assume that these open sets you mentioned line up in such a way that $F^{-1}\circ f \circ G (x_1,\ldots,x_m)=(x_1,\ldots ,x_n)$ since $m\geq n$. Then all you have to do is notice that the open set $W$ was open in euclidean space, while $U$ was an open nhood of a boundary point in the half-space. Personally I would say you have reached your contradiction. If you want to be more thorough, you can look at the last coordinate of your points in $U$ and notice they are all positive... I will leave that to you.

  • $\begingroup$ Hi N. Owad, thanks for your help. If $z$ is a point close to $x$, then, by continuity, $f(z)$ is close to $y$. Also, $dim f^{-1}(y)=m-n$, but i do not see how it can help. $\endgroup$
    – Framate
    Mar 30, 2014 at 22:18
  • $\begingroup$ @Miguemate So, an open neighborhood of $x$ is mapped to an open neighborhood of $f(x)$. But what do we know about $f(x)$? $\endgroup$
    – N. Owad
    Mar 30, 2014 at 23:33
  • $\begingroup$ $f(x)=y$, $y\in \partial Y$ a regular value $\endgroup$
    – Framate
    Mar 30, 2014 at 23:44
  • $\begingroup$ And what do open nhoods of boundary points look like? $\endgroup$
    – N. Owad
    Mar 31, 2014 at 0:00
  • $\begingroup$ I have this. For every $\epsilon>0$, there exists $\delta>0$ such that $y\in f(B_{\delta}(x))\subset B_{\epsilon}(y)=V_y\cap \partial Y$, $V_y$ an open subset of $\mathbb R^l$, assuming $Y\subset\mathbb R^l$. $\endgroup$
    – Framate
    Mar 31, 2014 at 0:24

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