Pullback of differential form under isotopic maps

Question

There's a smooth map $f$ from manifold $M$ to itself which is isotopic to identity (without changing the differential structure). Is there any conclusion we can draw on the form $f^*\tau-\tau$ where $\tau$ is a form on the manifold $M$? Can it be exact?

Answer 1

Without further hypotheses, no. For example, consider $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 2x$. This map is isotopic to the identity, and $f^*\tau - \tau = 2\tau - \tau = \tau$.

If you assume that $\tau$ is closed, then $f^*\tau - \tau$ is closed and in de Rham cohomology we have

$$[f^*\tau - \tau] = f^*[\tau] - [\tau] = \operatorname{id}^*[\tau] - [\tau] = [\tau] - [\tau] = 0.$$

That is, $f^*\tau - \tau$ is exact.