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Let $U,V$ be two open sets in $R^n$ and $f:U\to V$ proper $C^{\infty}$ map (proper = preimage of compact set is compact). Then we have

$$\int f^{*}\omega=\deg(f)\int \omega,$$

for $\omega \in \Omega_c^{n}(V)$. How to prove that if $f$ is linear mapping i.e. $f(x)=Ax$ for nonsignular $n\times n$ matrix we have $\deg(A)=sign(\det (A))$?

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I don't think that your property is true : we can show with Sard's theorem that the degree of a proper map is an integer (whereas your determinant could be a non-integer number). –  JBC Jun 16 '12 at 12:33
    
Yes, I have corrected it. Sorry! –  dmm Jun 16 '12 at 13:16

1 Answer 1

The degree of a linear map is the signum of $\det$, not $\det$ (the degree is integer valued, while $\det$ is real valued -- already this should make you suspicious). (And the answer to your question is, basically, the change of variables formula for the integral, once you replace $\det$ by it's sign).

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First thanks for your replies and sorry for my mistake. It should be sign of determinant rather that the determinant itself. I corrected it. Thomas, I need to use this lemma in order to derive change of variables, is there a way proving it without it? Thank you! –  dmm Jun 16 '12 at 13:15
    
@dmm If you post such a question it is usually hard to answer cause the people reading it do not know how the concepts were introduced to you and what you have available as tools. Maybe you have a standard textbook you are following, then those who know it may be able to help you based on this knowledge. The approaches to differential forms and degree I know all rely on the transformation formula (which reads $\int_{\phi(U)}f(x) d^nx = \int_U f(\phi(x))|\det D\phi(x)|d^n x$, just to make sure we are talking about the same thing). –  user20266 Jun 16 '12 at 13:43

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