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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! :)2011-03-31
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    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.2011-03-31
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    Good point, Mariano; information added.2011-03-31
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    And thank you for your answer! But I'm not too familiar with Algebraic Topology. Would you happen to know of any other explanations?2011-03-31
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    Ah! That was my initial attempt, but I had trouble creating the homotopy. Thank you!2011-04-01

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