Let $M,N$ be $n$-dimensional oriented smooth manfiolds without boundary and $\omega$ a compactly supported $n$-form on $N$. If $f \colon M \rightarrow N$ is an orientation-preserving diffeomorphism, then $$ \tag{1} \label{a} \int_{N}\omega = \int_{M}f^{\ast}\omega. $$

I want to prove (\ref{a}) using the change of variables formula in $\mathbb{R}^n$, which says that if $\psi \colon U \rightarrow V$ is an orientation-preserving diffeomorphism between open subsets of $\mathbb{R}^n$, then $$ \tag{2} \label{b} \int_{V}\eta = \int_{U}\psi^{\ast}\eta, $$ where $\eta$ is an $n$-form having compact support in $V$.

Try of proof: W.l.o.g. we can assume (otherwise take a partition of unity) that there is an orientation-preserving chart $\varphi \colon U \rightarrow O$ such that $\operatorname{supp} \omega \subseteq O$, where $O \subseteq N$ is open. Therefore by definition of the integral $$ \tag{3} \label{c} \int_{N}\omega = \int_{U} \varphi^{\ast}\omega. $$ On the other hand, $\operatorname{supp}f^{\ast}\omega \subset f^{-1}(O)$ and $f^{-1} \circ \varphi$ is a chart for $M$ with $\operatorname{supp}f^{\ast}\omega \subset \operatorname{im} f^{-1} \circ \varphi$. Therefore $$ \tag{4} \label{d} \int_{M}f^{\ast}\omega = \int_{U}(f^{-1} \circ \varphi)^{\ast}f^{\ast}\omega. $$ At this point I would be tempted to write $(f^{-1} \circ \varphi)^{\ast}f^{\ast}\omega = \varphi^{\ast}\omega$ and say the integrals (\ref{c}) and (\ref{d}) are equal, but I think something is wrong in this argument. However, how can I rewrite (\ref{a}) to reduce it to the case (\ref{b})?

  • $\begingroup$ Note that in your "by definition of the integral" you need to use the fact that the chart is itself orientation-preserving. $\endgroup$ – Ted Shifrin May 28 '17 at 15:47
  • $\begingroup$ @TedShifrin Thank you, I corrected it. $\endgroup$ – user98187609 May 28 '17 at 15:48
  • $\begingroup$ Did you correct it both places? That's what is missing to make your argument correct. $\endgroup$ – Ted Shifrin May 28 '17 at 15:49
  • $\begingroup$ I think in the second place we have that $f^{-1}$ is orientation-preserving (by assumption) and $\varphi$ is (still) orientation-preserving, therefore $f^{-1} \circ \varphi$ is orientation preserving. $\endgroup$ – user98187609 May 28 '17 at 15:52
  • $\begingroup$ I am a bit confused. Actually, I am trying to understand the proof of (1) which can be found in "Introduction to Smooth Manifolds" by Lee and there it just says that (1) follows from (2) after choosing a chart like I did. (and the definition of the integral) But it looks like I did not use (2). $\endgroup$ – user98187609 May 28 '17 at 15:56

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