Integrals of Pullbacks

Question

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?

Answer 1

Following the hint, if $F: \partial W \to Y$ is a smooth map that extends to all $W$ and $\omega$ is a closed $k$-form on $Y$, then $d\omega=0$, thus Stokes Theorem gives $$ \int_{\partial W} F^*\omega = \int_{W} d(F^*\omega) =\int_{W} F^*(d\omega) = \int_W 0 = 0 \tag{1} $$

Now let $F: X \times I \to Y$ be an homotopy between $f_0$ and $f$. Since $X\times I$ is a boundary manifold, with boundary given by $X\times \{ 0 \} \cup X \times \{ 1 \}$, where $\{ 0\}$ has orientation $-1$ and the orientation of $\{ 1 \}$ is $+1$, then $$ \int_{\partial(X \times I)} F^* \omega = \int_{ X \times \{ 1 \}} F^*\omega - \int_{ X \times \{ 0 \}} F^*\omega = \int_{X} f_1^* \omega -\int_{X} f_0^* \omega $$ However by (1), it follows that $\int_{\partial(X \times I)} F^* \omega=0$, hence the claim follows.

Answer 2

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.