The following is adapted from my professor's summary notes on Differential Geometry:
Let $M$ and $N$ be two oriented connected manifolds of the same dimension $m$, and let $f \colon M \to N$ be a proper map. Let $I \colon H_{\mathrm{c}}^m(N) \to \mathbf{R}$ be the isomorphism given by integration from the $m^{\mathrm{th}}$-degree compactly supported de Rahm cohomology space of $N$ to the real numbers, and similarly consider $I \colon H_{\mathrm{c}}^m(M) \to \mathbf{R}$, as well as the map on cohomology $H_\mathrm{c}^m(f) \colon H_\mathrm{c}^m(N) \to H_\mathrm{c}^m(M)$ induced by $f$.
Then the linear map $I \circ H_\mathrm{c}^m(f) \circ I^{-1} \colon \mathbf{R} \to \mathbf{R}$ is multiplication by a real number $\mathrm{deg}(f)$ which we call the cohomological degree of the proper map $f \colon M \to N$. ....
He then continues to define the geometric degree $\mathrm{deg}_y(f)$ of $f$ at a regular value $y \in N$ in the usual way as the sum of local intersection numbers. He then states:
Theorem. The geometric degree $\mathrm{deg}_y(f)$ of $f \colon M \to N$ at each regular value $y \in N$ is equal to the cohomological degree $\mathrm{deg}(f)$.
....[some corollaries]
Corollary. If $M$ is compact and $N$ is non-compact, the degree $\mathrm{deg}(f)$ is $0$.
Again, these notes have statements only, no proofs.
Could someone help me understand why the corollary is true? I really have no idea. Thanks.
The map in that case is not surjective. Compute the geometric degree at a point which is not in the image.