# The mod 2 degree of a function when the image space N has a boundary

I was flipping through Milnor's "Topology from the Differentiable Viewpoint," and I came upon a sentence concerning the mod 2 degree of a function from M to N. It essentially says: "We may as well assume also that N is compact without boundary, for otherwise the deg mod 2 would necessarily be zero."

I understand why this is true for compactness. However, any ideas I have trying to prove the boundary part seem way to complex for a paranthetical aside. Any ideas are appreciated.

Edit: Also, M and N are smooth manifolds of the same dimension, f is smooth, and M is assumed to be compact and boundaryless (so the definition of mod 2 degree makes sense).

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We can guess that $M$ and $N$ are smooth manifolds, that the map is smooth, &c But you can also add that information to the body of your question! :) –  Mariano Suárez-Alvarez Mar 31 '11 at 17:05
If $N$ has dimension $n$ and is $R$-oriented, Lefschetz duality gives you an isomorphism $H^n(N; R)\cong H_0(N,\partial N; R)$; if $\partial N$ intersects every path-component of $N$, the latter group is zero. –  Mariano Suárez-Alvarez Mar 31 '11 at 17:16
Good point, Mariano; information added. –  user8946 Mar 31 '11 at 17:25
And thank you for your answer! But I'm not too familiar with Algebraic Topology. Would you happen to know of any other explanations? –  user8946 Mar 31 '11 at 17:26
Ah! That was my initial attempt, but I had trouble creating the homotopy. Thank you! –  user8967 Apr 1 '11 at 0:53

One way to explain this is observing that Lefschetz duality gives you, when $N$ is $R$-orientable (for example, if $R=\mathbb Z/2\mathbb Z$) an isomorphism $H^n(N; R)\cong H_0(N,\partial N; R)$ and that if $\partial N$ intersects every path-component of $N$ then the latter group is zero.
Alternatively, if $f:M\to N$ is a map between $n$-manifolds and $N$ has a non-empty boundary, you can show that $f$ is homotopic to a map which is not surjective and therefore of degree zero: if $B$ is a connected component of $\partial N$, then $B$ has a neighborhood diffeomorphic to $B\times[0,1)$, and then you can deform $f$ by sliding it along the $[0,1)$ factor into another function $g:M\to N$ which misses all of $B\times[0,\tfrac12)$.