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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...

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A colleague suggested the following: Suppose that $N$ is a compact manifold with boundary. Then, sew $N$ with itself along the boundary, call it $\hat{N}$. So now $f$ can be thought of as a non-surjective map from $M$ into a compact manifold without boundary $\hat{N}$, so the degree of $f$ is 0 mod 2... –  Braindead May 25 '12 at 21:19
    
I realized that this is a rather trivial question: any surjective map onto a manifold with boundary can be deformed by homotopy to a non-surjective map. –  Braindead Jun 4 '12 at 20:09

1 Answer 1

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

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