[Math] Mapping Degree of a Smooth Map from a Compact Manifold without Boundary

Question

There is a comment in Milnor's "Topology from a Differentiable Viewpoint," that I don't quite understand:

Let $f$ be a smooth map from $M$ to $N$, where $M$ is compact without boundary, and $N$ is connected, and both manifolds have the same dimension. (We may as well assume also that $N$ is compact without boundary, since otherwise the mod 2 degree would necessarily be zero.)

So if $N$ is non-compact, then $f$ cannot be surjective, so I see that the mod 2 degree would be zero.

Why does $N$ being compact with boundary would give 0 mod 2 degree? The examples I can think of all have 0 mod 2 degree, but I'm not sure why this is obvious…

Answer 1

Any surjective map onto a manifold with boundary is homotopic to a non-surjective map, by "rolling up the sleeves."