# $\int _c \omega$ is independent of orientation-preserving re-parameterization of c

I'm working on the following problem from Guillemin and Pollack's Differential Topology:

Let $c : \left[a, b\right] \rightarrow X$ be a smooth curve, and let $f: \left[a_1, b_1\right] \rightarrow \left[a, b\right]$ be a smooth map with $f(a_1) = a$ and $f(b_1) = b$. Show that the integrals

$$\int _a ^b c^* \omega \text{ and } \int _{a_1} ^ {b_1} (c \circ f)^* \omega$$ are the same.

While it doesn't say so explicitly, I'm assuming $\omega$ is a compactly supported smooth $k$-form on the $k$-dimensional manifold $X$.

I know that if $c$ and $f$ are both orientation-preserving diffeomorphisms, then using a theorem in Guillemin and Pollack's book

$$\int _{a_1} ^{b_1} (c \circ f)^*\omega = \int _{a_1} ^{b_1} f^* c^* \omega = \int _a ^b c^* \omega = \int _c \omega$$

However, I don't really see any reason to believe $c$ and $f$ are both orientation-preserving diffeomorphisms, so I'm not sure how to proceed. I know $c$ is a smooth map onto its image, but I can't see why it should be one-to-one. Can anyone provide some advice, please? In case it's relevant, this is Exercise 4 from Section 4.4 of Guillemin and Pollack's book, and this problem is for my own self study, not from a graded assignment.

$\omega$ should be a smooth $1$-form as you want to integrate this along a curve $c$. $\omega$ doesn't have to be of compact support, as $c$ is compact anyway.
Indeed, after the pullback $c:[a, b]\to X$, the calculations are done on $[a,b]$. Write $c^* \omega = g(x) dx$ on $[a,b]$. Then $$(c\circ f)^* \omega = f^* c^* \omega = f^* (g(x) dx) = g(f(x)) f'(x) dx.$$ So you are asking why $$\int_{a_1}^{b_1} g(f(x)) f'(x) dx = \int_a^b g(x) dx.$$ But this is true as a rule of substiution, which folllows from Chain rule. From the link you can also see how $f(a_1) = a$ and $f(b_1) = b$ are used.
• Thank you for your answer. I just have one quick question. In my textbook it seems like in the definition of the integral of a differential form over a manifold it requires that the form is compactly supported. Why doesn't $\omega$ have to be of compact support here? – Gecko Nov 23 '15 at 3:37
• If your manifold is $n$-dimensional and $\omega$ is a $n$-form, then you require that $\omega$ is of compact support. Now you are integrating $c^*\omega$ over $[a,b]$, so $c^*\omega$ has compact support on $[a,b]$ anyway. @Gecko – user99914 Nov 23 '15 at 3:54