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I'm just a little stuck at the end of proving this. I have seen other similar questions but they don't address where I am stuck.

Let $\mu$ be an orientation form over $M$, and equip $M$ with the orientation enduced by $\mu$. Choose an orientation-preserving atlas $\{U_i,\varphi_i\}$ and a partition of unity $\{\theta_i\}$ subordinate to this atlas. Then $\int_M=\sum_i\int_M\theta_i\mu$. Consider the $i$th term in the sum. Since the support of $\theta_i$ is contained in $U_i$, then $$\int_M\theta_i\mu=\int_{U_i}\theta_i\mu=\int_{\varphi_i(U_i)}(\varphi_i^{-1})^*\theta_i\mu$$

Now, I am not sure why the last integral is positive.

trystero
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    I think this should be guaranteed by the definition of orientation-preserving atlas. What is that definition? – Matthew Leingang Dec 15 '16 at 14:08
  • You are right, it does follow from the definition. In particular, an orientation-preserving map sends positively oriented forms to positively oriented forms. Since $\theta_i$ is non-negative then $\theta_i\mu$ is positively oriented where it is not identically zero. Then the pullback of the from to $\mathbb{R}^n$ is positively oriented, so it is some positive function times the standard $n$-form on $\mathbb{R}^n$. Then the integral is positive. Thanks for the help. – trystero Dec 15 '16 at 15:36
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    Glad you figured it out. Just to make sure this question has an answer, I suggest you make that comment your answer and accept it. – Matthew Leingang Dec 15 '16 at 16:18

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It follows from the definition. In particular, an orientation-preserving map sends positively oriented forms to positively oriented forms. Since $\theta_i$ is non-negative then $\theta_i\mu$ is positively oriented where it is not identically zero. Then the pullback of the form to $\mathbb{R}^n$ is positively oriented, so it is some positive function times the standard $n$-form on $\mathbb{R}^n$. Then the integral is positive.

trystero
  • 701