TITLE: Integrals of Pullbacks
QUESTION [6 upvotes]: This is a problem from Guillemin's Differential Topology:

Suppose that $f_0, f_1: X \to Y$ are homotopic maps and that the compact boundaryless manifold $X$ has dimension $k$. Prove that for
  all closed $k$-forms $\omega$ on $Y$, $$ \int_X f_0^*\omega = \int_X
f_1^*\omega.$$

There is a hint to use a result from a previous exercise, which says that if $f:\partial W \to Y$ is a smooth map that extends to all of $W$ and $\omega$ is a closed $k$-form on $Y$, with $k = \dim \partial W$, then 
$$\int_{\partial W} f^*\omega = 0.$$
It seems to me this should be a straightforward exercise, but I can't get it done. Any hints?

REPLY [2 votes]: We need to use the Natural Transformation Law defined in Guillemin p.168 and p.176:

If $f:Y\to X$ is an orientation-preserving diffeomorphism, then $\int_X{\omega}=\int_Y{f^*\omega}$ for every compactly supported, smooth $k$-form on $X$.

We can now claim  $\int_X{f^*_j\omega}=\int_Y{\omega}$, where $j=0$ or $1$, which would be enough, but we observe that $\omega$ is closed in $Y$, and it's pullback might be bounded in $X$.
So $\int_X{f^*_j\omega}$ is misleading, and we should write $\int_{X \cup \partial X}{f^*_j\omega}$. But this means we have $\int_{X \cup \partial X}{f^*_j\omega}=\int_X{f^*_j\omega}+\int_{\partial X}{f^*_j\omega}$, and the second integral on the RHS is zero, so the result holds.