Homotopy invariance of de Rham cohomology Let $M,N$ be smooth manifolds which are homotopy equivalent i.e., there exists smooth maps $F:M\rightarrow N$ and $G:N\rightarrow M$ such that $F\circ G$ is homotopic to identity map on $N$ and $G\circ F$ is homotopic to identity map on $M$.
Then, Homotopy invariance of deRham cohomology says that the de Rham cohomology groups of $M$ and $N$ are isomorphic.
I am not able to understand the construction given in Lee's Intoduction to Smooth manifolds. 
What is the rough idea behind this proof (or any other proof) of homotopy invariance of de Rham cohomology.
EDIT : Given that $M,N$ are homotopy equivalent as above, we need to prove that $H^p_{dR}(M)$ and $H^p_{dR}(N)$ are isomorphic. we expect this to come from $F^*:H^p_{dR}(N)\rightarrow H^p_{dR}(M)$ and $G^*:H^p_{dR}(M)\rightarrow H^p_{dR}(N)$.
I do not understand the idea behind proof of two homotopic maps have induce same deRham cohomology maps.
Once we prove this, then $F\circ G$ and $1_N$ induce same deRham cohomology maps i.e., the composition $H^p_{dR}(N)\xrightarrow{F^*} H^p_{dR}(M)\xrightarrow{G^*} H^p_{dR}(N)$ is same as the identity map on $H^p_{dR}(N)$ and similarly the composition $H^p_{dR}(M)\xrightarrow{G^*} H^p_{dR}(N)\xrightarrow{F^*} H^p_{dR}(M)$ is same as the identity map on $H^p_{dR}(M)$. This says that $F^*\circ G^*=1$ and $G^*\circ F^*=1$. Thus,  $F^*,G^*$ are isomorphisms, inverses to each other, conlcuding that deRham cohomology groups  $H^p_{dR}(M)$ and $H^p_{dR}(N)$ are isomorphic.
How do we prove that two homotopy maps induce same deRham cohomology maps. Let $f:M\rightarrow N$ and $g:M\rightarrow N$ be two homotopy maps, we want to prove that $f^*=g^*:H^p_{dR}(N)\rightarrow H^p_{dR}(M)$ i.e., $f^*(\omega)=g^*(\omega)+\text{closed p-form on }M$ when seen as maps $\Omega^p(N)\rightarrow \Omega^p(M)$. This means, we are expected to have $$f^*(\omega)=g^*(\omega)+d\eta$$ where $\eta$ is a smooth $p-1$ form. 
This gives question of defining a map $h:\{\text{closed p-forms on }N\}\subseteq \Omega^p(N)\rightarrow \Omega^{p-1}(M)$ assigning to each closed $p$ form $\omega$ on $N$ a $p-1$ form $\eta$ on $M$ such that  $f^*(\omega)=g^*(\omega)+d\eta$.
Then author says it turns out to be far simpler to define $h:\Omega^p(N)\rightarrow \Omega^{p-1}(M)$ not with the condition $f^*(\omega)=g^*(\omega)+d(h\omega)$ for every closed form $\omega$ but with a more general condition that $$f^*(\omega)-g^*(\omega)=d(h\omega)+h(d\omega)$$
for every smooth $p$ form. Suppose $\omega$ is closed then $d\omega=0$ and we get the required condition that $f^*(\omega)-g^*(\omega)=d(h\omega)$.
So, now the question is to define a map $h:\Omega^p(N)\rightarrow \Omega^{p-1}(M)$ satisfying the condition as above. How can we think of constructing such map? If we are thinking of going from a $p$ form to a $p-1$ form one obvious thing is to some how integrate this $p$ form. What $p$ form can we integrate here? It is natural to some how integrate the $p$ form $f^*(\omega)-g^*(\omega)$ to get a $p-1$ form $h\omega$. So, when you reverse the process i.e., when you differentiate you get $f^*(\omega)-g^*(\omega)=d(h\omega)$. This idea is vague and I can not make it any better.
This $h$ is called a homotopy operator in this book.
Any suggestions on how would you think about producing this operator is welcome.
 A: 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.
