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.