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I'm attempting Exercise 8.21 from Spivak's Differential Geometry. It is not for homework or anything. The problem states

Let $f\colon M^n\to N^n$ be a proper map between oriented $n$-manifolds such that $f_*\colon M_p\to N_{f(p)}$ ($f_*$ here is $df_p$ in other texts I think) is orientation preserving whenever $p$ is a regular point. If $N$ is connected, then either $f$ is onto $N$, or else all points are critical points of $f$.

First, since $f$ is a proper map between manifolds, it is closed, and thus $\operatorname{im}(f)\subset N$ is closed. I want to show, assuming that there is a regular point of $f$ in $M$, then $\operatorname{im}(f)$ is open, hence all of $N$ since $N$ is connected.

If $p$ is a regular point, so $f_*\colon M_p\to N_{f(p)}$ is surjective, so is also bijective since $M$ and $N$ are both $n$-dimensional. By the inverse function theorem, $f$ is a local diffeomorphism, so maps an open neighborhood of $p$ onto an open neighborhood of $f(p)$. If I could show arbitrary $q\in\operatorname{im}(f)$ is in one of these neighborhoods, then I would be done.

If $q$ is not in the diffeomorphic image of any open neighborhood of a regular point of $f$, does this contradiction the orientation preserving property somehow?

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This is a mapping degree matter. As $N$ is connected, the Brouwer-Kronecker degree is well defined. One picks any regular value $q\in N$ and look at the sum $$ \deg(f)=\sum_{f(p)=q}sign\, d_pf, $$ where $sign$ is $+1$ if $d_p$ preserves orientation, $-1$ otherwise. Any $q\notin im(f)$ is trivially a regular value and the sum is of course null. But if there is some regular value $q\in im(f)$ with $sign\, d_pf=+1$ for all $f(p)=q$ then the sum cannot be zero! Hence the conclusion.

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  • $\begingroup$ Does this argument work in case of manifolds with boundary? $\endgroup$ – Peter Franek Mar 13 '15 at 19:32
  • $\begingroup$ There is indeed a notion of mapping degree for manifolds with boundary, but rather than Brouwer-Kronecker it is Euclidean type. In that case homotopies doesn't touch the boundary... Of course there are connections among the two approaches. But I cannot be categorical here. I'll think of it! $\endgroup$ – Jesus RS Mar 13 '15 at 20:05
  • $\begingroup$ (@Peter Franek: sorry for the delay) Without conditions on the boundary there is trouble. Think of the inclusion $M\hookrightarrow\mathbb S^2$ from the upper closed semisphere into the sphere. $\endgroup$ – Jesus RS Mar 24 '15 at 14:30

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