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We'll use the Proposition (F) to show that:

(Invariance of domain) Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected. If $f$ is injective then $f$ is a open map.

dem: We can assume that $f$ is locally proper by $f:M→Y$ $C^{1}$ map between manifolds $M$ e $N$ with dimension m and n respectively then $f$ is locally proper.

Let $x \in M$, and $y:=f(x) \in N$ with $(U,\phi)$ chart in $M$ of $x$ and $(V,\psi)$ chart in $N$ of $f(x)$ such that $f(U)\subseteq V$ is proper. Let $B(\psi(y), \epsilon) \subseteq \psi(V)$ so $\psi^{-1}(B(\psi(y), \epsilon)$ is a open connected subset of $N$. Since $f$ is continous, exist $W$ open in $M$ such that $W \subseteq U$ and $f(W) \subseteq \psi^{-1}(B(\phi(x), \epsilon)$.

Let $B(\phi(x), \delta) \subseteq \phi(W)$ so $\phi^{-1}(B(\phi(x), \delta)$ is a open connected subset of $M$ such that

$f(\phi^{-1}(B(\phi(x), \delta)) \subseteq \psi^{-1}(B(\psi(y), \epsilon)$.

Note that $\tilde{f} = f|_{\phi^{-1}(B(\phi(x), \delta)}: \phi^{-1}(B(\phi(x), \delta) \to \psi^{-1}(B(\psi(y), \epsilon)$ is injective proper mapping of two oriented, boundaryless, smooth manifolds of dimension $m$, where where $\psi^{-1}(B(\psi(y), \epsilon)$ is connected. Then by (F) $\tilde{f}$ is surjective. So $f( \phi^{-1}(B(\phi(x), \delta))$ is open, and we finished.

In this solution has a problem: I'd take $\psi^{-1}(B(\psi(y), \epsilon + \xi)$ and the result continous true, so $\tilde{f}$ is a bijecion of $\phi^{-1}(B(\phi(x), \delta)$ to $\psi^{-1}(B(\psi(y), \epsilon)$ and $\psi^{-1}(B(\psi(y), \epsilon +\xi)$, and it's false.

Where am I missing???

Thank you.


We have that:

Definition (A):

Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; and $x$ is a regular point of $f$. So $df_{x}$ is a linear isomorphism. If $f$ preserves (resp. reverse) orientation at $x$ we denote $sing_{x}(f) = +1$ (resp. $-1$).

(Mapping Degree, Ruiz, 2009. p. 86)

Proposition (B):

Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; and $y$ is a regular value of $f$. so $f^{-1}(y)$ is a finite set and $d(f,y)= \sum_{x \in f^{-1}(y)} sign(df_{x})$.

(Mapping Degree, Ruiz, 2009. p. 95)

Proposition (C):

Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected. Then the integer $d(f,y)$ does not depend on the choice of regular valeu $y$. It is called the degree of $f$ an is denoted by $deg(f)$.

(Mapping Degree, Ruiz, 2009. p. 99)

Corollary (D):

Let $f: M \to N$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected. If the $deg(f) \neq 0$ then $f$ is surjective map.

(Mapping Degree, Ruiz, 2009. p. 99)

Proposition (E):

If $f: U \to \mathbb{R}^{n}$ $C^{1}$ locally injective and $U$ is a open set in $\mathbb{R}^{n}$ then the set of the points $x \in U$ such that $df_{x}$ is injective is open and dense in $U$.

dem: By $rank$ $theorem$.

Proposition (F):

Let $f: M \to N$ $C^{1}$ be a proper smooth mapping of two oriented, boundaryless, smooth manifolds of dimension $m$; furthermore, $N$ is connected. If $f$ is injective then $f$ is surjective.

dem: Let $(U,\phi)$ chart in $M$. By (E) there exists $x \in U$ such that $df_{x}$ is injective, so $df_{x}$ is a isomorphism linear. So $y=f(x)$ is regular value, and since $f$ is injective we have that $d(f,y) \neq 0$. So by (D), we have that $f$ is surjective.

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Your proof of invariance of domain doesn't work, because you have given no justification that $\tilde{f}$ is proper. In fact, it probably won't be proper. Indeed, if (as you suggest) you replace $\epsilon$ by $\epsilon+\xi$ for some $\xi>0$, then $\tilde{f}$ will definitely not be proper, since $\psi^{-1}(\overline{B(\psi(y), \epsilon)})$ is a compact subset of $\psi^{-1}(B(\psi(y), \epsilon+\xi))$ containing the image of $\tilde{f}$, but the domain of $\tilde{f}$ is not compact.

To fix the proof, you need to be much more careful in choosing the domain of $\tilde{f}$. You can take the domain of $\tilde{f}$ to be $f^{-1}(\psi^{-1}(B(\psi(y), \epsilon)))$: then $\tilde{f}$ will be proper because $f$ was. (Properness of a map is preserved if you restrict the range to a subset $A$ and the domain to $f^{-1}(A)$.) To conclude that $\tilde{f}$ (and hence $f$) is open, you then need to use the fact that a proper continuous bijection between locally compact Hausdorff spaces is a homeomorphism.

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  • $\begingroup$ Hi. About the first paragraph, $\tilde{f} = f|_{\phi^{-1}(B(\phi(x), \delta)}: \phi^{-1}(B(\phi(x), \delta) \to \psi^{-1}(B(\psi(y), \epsilon) $ is proper. Let $K$ compact in $\psi^{-1}(B(\psi(y), \epsilon)$, so $K$ is closed. By continuous of $\tilde{f}$, we have that $\tilde{f}^{-1}(K)$ is closed. So $\phi(\tilde{f}^{-1}(K))$ is closed, but $\phi(\tilde{f}^{-1}(K)) \subseteq B(\phi(x), \delta)$ bounded set, so $\phi(\tilde{f}^{-1}(K))$ is compact, then $\tilde{f}^{-1}(K)$ is compact, and $\tilde{f}$ is proper. $\endgroup$ – Alladin Apr 21 '16 at 18:01
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    $\begingroup$ You know that $\phi(\tilde{f}^{-1}(K)) \subseteq B(\phi(x), \delta)$ is closed in $B(\phi(x), \delta)$, but why is it closed in all of $\mathbb{R}^m$? You need to know that in order to conclude it is compact. $\endgroup$ – Eric Wofsey Apr 21 '16 at 18:04
  • $\begingroup$ In last paragraph to conclude that $\tilde{f}$ is open can we change the fact that "a proper continuous bijection between locally compact Hausdorff spaces is a homeomorphism" by proposition (F) ? $\endgroup$ – Alladin Apr 21 '16 at 18:14
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    $\begingroup$ No, proposition (F) is what gives you that $\tilde{f}$ is a bijection, but you still have to do some additional work to conclude it is a homeomorphism. $\endgroup$ – Eric Wofsey Apr 21 '16 at 18:15
  • $\begingroup$ Hi, I found a result due Smale that showed that $f$ is locally proper. math.stackexchange.com/questions/1753360/… $\endgroup$ – Alladin Apr 21 '16 at 21:20

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