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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?

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    $\begingroup$ A homotopy gives you a map $f:X\times I\rightarrow Y$ where the boundary is $X\sqcup X$. $\endgroup$
    – J126
    Jun 14, 2015 at 17:35
  • $\begingroup$ Isn't the symbol $\sqcup$ used to denote disjoint unions? $\endgroup$
    – Douglas Finamore
    Jun 14, 2015 at 18:26
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    $\begingroup$ Yes, which is what Joe Johnson is denoting. More precisely the boundary is $X \sqcup -X$, where the latter means $X$ with the opposite orientation. $\endgroup$
    – user98602
    Jun 14, 2015 at 21:58
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    $\begingroup$ As Joe Johnson said, the homotopy is a map,f:X×I→Y. $f^*(\omega)$ is a closed form on X×I. By Stokes theorem its integral over the boundary is zero. But the integral over the boundary is the difference of the integrals of $f^*_0(\omega)$ and $f^*_1(\omega)$ over X. $\endgroup$
    – Joe S
    Jun 20, 2015 at 14:50
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    $\begingroup$ @DouglasFinamore are you still looking for an answer? It seems like these comments cover it. Do you agree? $\endgroup$
    – James S. Cook
    Jun 21, 2015 at 5:04

2 Answers 2

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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.

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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.

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    $\begingroup$ @LeoSera: The answerer has written $X + dX$ instead of $X \cup \partial X$. I shall correct this, but I think the idea is clear. Next, as the answerer clearly writes, $k$ in $f_k$ stands for $0$ or $1$, read the answer again carefully. You might want to reconsider the downvote. $\endgroup$
    – Alex M.
    Jun 23, 2015 at 19:56
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    $\begingroup$ @AlexM. Oh I see, but anyway bad choice putting $k=0,1$ since $k$ is the dimension of the manifold. Still pretty confusing. Also the question had already a clear answer so I am keeping the downvote ! $\endgroup$
    – Leo Sera
    Jun 23, 2015 at 21:48

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