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How can I prove the following statement?

Let $f:M\rightarrow N$ be a smooth map between closed, connected, oriented manifolds of the same dimension. If $\deg(f)\neq 0$, then $f$ is surjective.

Here $\deg(f)$ denotes the Brower degree of $f$.

Note: This is not homework, but a question that came up while studying for a course of differential geometry. I tried to prove the statement in more than a way, but always noticed that what I did usually led to nowhere. Any hint or insights would be greatly appreciated.

Daniel Robert-Nicoud
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    Take a point not in the image of $f$. Is that a regular value? – Ryan Budney Aug 02 '13 at 21:08
  • @RyanBudney Oh right, that's true! Thus $# f^{-1}(x)=0$ and that would be a contradiction, right? But still there is a detail that bothers me: consider the identity map of the interval $[0,1]$. It has degree $1$. But it is homotopic to a constant map, which should have degree $0$. This would be a contradiction, so where am I doing wrong? Anyway write down your comment as an answer, and I will accept it. Thank you. – Daniel Robert-Nicoud Aug 02 '13 at 21:13
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    You've contradicted your assumption that the manifold is closed. The degree isn't even defined for maps of manifolds with boundary. – Ryan Budney Aug 02 '13 at 21:14
  • @RyanBudney Ah, I see. Is the problem that you can't show invariance under homotopies, since if $M$ is a manifold with boundary then $M\times[0,1]$ is not a manifold? And is the implication $f$ surjective $\rightarrow\deg(f)\neq 0$ true? – Daniel Robert-Nicoud Aug 02 '13 at 21:18
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    @Daniel: There are plenty of surjective maps which have $0$ degree. For example, consider the map of a circle which maps the top half all the way around once and the bottom half arond the opposite way. – Jason DeVito - on hiatus Aug 03 '13 at 00:25

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