If $f\colon M^n\to N^n$ is proper, $N$ is connected, and $f_*$ preserves orientation at regular points, then $f$ is surjective? 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?

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.
