Homotopy invariance of de Rham cohomology

Now I read the book and realize that Lie derivative is introduced after the chapter on cohomology, if the order is reversed there is a very direct interpretation.

You want to prove:

If $f_0,f_1 : M\to N$ are smooth mapping which are homotopic, then $$ f_0^* = f_1^* : H^k_{dR}(N)\to H^k_{dR}(M)$$ for all $k$.

Recall that the induced pullback mapping on $H^k$ is just $f_0^* [\alpha] = [f_0^*\alpha]$ and similar for $f_1$. So you need to show: for any $k$-form $\alpha$ on $N$, $ [f_0^*\alpha] =[ f_1^*\alpha]$, or $[f_1^*\alpha - f_0^*\alpha] = 0$.

That is, you want to write $f_1^*\alpha - f_0^*\alpha$ as $d$ of something. Note that by the fundamental theorem of calculus,

$$f_1^*\alpha - f_0^*\alpha = \int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt.$$

Here $f_t$ is the homotopy between $f_0$ and $f_1$. Of course it is not clear what the right hand side is. We want to give it a more intrinsic interpretation, so that we can check if the right hand side is really $d$ of something.

We let $F : M \times [0,1] \to N$ be the homotopy and $\iota_t : M\to M\times [0,1]$, $\iota_t (x) = (x, t)$ be the inclusion. Then $f_t = F\circ \iota_t$, thus $f_t^* = \iota_t^* \circ F^*$ and

$$\begin{split} \frac{\partial}{\partial t} (f_t^* \alpha) (x)&= \frac{\partial }{\partial t} (\iota_t^* (F^*\alpha) (x)) \\ &= \frac{\partial}{\partial t} (F^*\alpha)(x, t)\\ &= \mathscr{L}_T (F^*\alpha), \end{split}$$ $\mathscr L_T$ is the Lie derivative along the vector $T:=\frac{\partial}{\partial t}$ (as a vector field on $M\times [0,1]$). Now the Cartan's magic formula gives (for any differential form $\omega$, vector fields $X$)

$$ \mathscr L_X \omega = \iota_X d\omega + d \iota_X \omega.$$

So we have

$$\begin{split} \int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt &= \int_0^1 \mathscr L_T (F^*\alpha) dt \\ &= \int_0^1 \big( \iota_T d(F^*\alpha) + d\iota_T (F^*\alpha)\big) dt \\ &= \int_0^1 \iota_T F^* (d\alpha) dt+ d \left( \int_0^1 \iota_T (F^*\alpha) dt\right) \end{split}$$

Note that the integration are exactly the homotopy operator $h$ constructed: so

$$ \int_0^1 \frac{\partial}{\partial t} (f_t^*\alpha) dt = h(d\alpha) + d(h(\alpha)).$$

So we have the next best thing: the right hand side in general is not $d$ of something, but it is when $\alpha$ is closed. This proves the theorem.

Of course I am just hiding everything in the Cartan's magic formula. The formula is commonly proved by direct calculation. A more fancy/geometric argument is suggested in Arnold's classical mechanics here. Note that the latter one also use a homotopy operator.