How to find or guess the homotopy operator? In the proof of the Poincare Lemma for compactly supported cohomology, the homotopy operator $K$ suddenly appears and satisfies the equation $1-eπ=\pm（dK-Kd)$. That is too lucky! I do not know how to find or guess what the $K$ is. Who can help me?
 A: Edit: In comparison to the first "draft", I have shortened it considerably. Which means that I have excluded all computations. 
I think we can say that one simply has to go through the trouble of actually computing the necessary conditions for such a homotopy operator to be able to guess it. --- No magic involved.
Since I have just started working through the book of Bott/Tu and I have asked myself this same question, I will give a shot at an answer. (I do not quite remember the proof of the Poincaré lemma for compact support at this point, so I'll try to figure it our from scratch...)
We want to show that $H^\ast_c(M \times \mathbb R) \simeq H^{\ast - 1}_c(M)$ for any manifold $M$. Certainly the natural maps between $M \times \mathbb R^n$ and $M$ are inclusion $i: M \to M\times \mathbb R$ and projection $\pi: M \times \mathbb R \to M$. They induce maps on forms by their pullbacks 
$$i^\ast: \Omega^\ast(M\times \mathbb R) \to \Omega^\ast(M),\quad \pi^\ast: \Omega^\ast(M) \to \Omega^\ast(M\times \mathbb R)$$
Now the problem that immediately arises is that the pullback of a compactly supported form will not be compactly supported for $\pi^\ast$. 
Suppose we are given a form $\phi = f(x) dx^I \in \Omega^k(M)$, then the pull-back $\pi^\ast \phi$  will have noncompact support in the vertical direction. So to remedy that, it seems natural to wegde it with a compactly supported, closed form $e$ in the vertical direction $\mathbb R$. We can of course take $e = e(t) \, dt$ to have total integral $\int_{\mathbb R} e(t) \, dt = 1$. Then (by identifying $e$ with the corresponding form on $M\times \mathbb R$) the form
$$\pi^\ast \phi \wedge e = f(x)e(t) \, dx^I \wedge dt \in \Omega^{k+1}(M\times \mathbb R)$$
has compact support and will be closed if $\phi$ is. This construction gives us a map
$$e_\ast: \Omega_c^\ast(M) \to \Omega_c^{\ast+1}(M\times \mathbb R), \quad e_\ast(\phi) = \left(\pi^\ast \phi\right) \wedge e$$
By construction, it is easy to see that this map must have a left-inverse, which is given by integrating along the fiber $\mathbb R$ (and denoted by $\pi_\ast$). So we know that $\pi_\ast \circ e_\ast$ is simply the identity (even on forms), we have to check that as a map in cohomology, $e_\ast \circ \pi_\ast$ is also the identity. For this, it suffices to show that on forms 
$$\phi - (e_\ast\circ \pi_\ast) \phi = K(d\phi) \pm dK(\phi)$$
where $K: \Omega_c^\ast(M\times \mathbb R) \to \Omega_c^{\ast - 1}(M \times \mathbb R)$ is a homotopy operator. 
Let us denote the coordinates on $M \times \mathbb R$ by $(x_1, \dots, x_n, t)$ in the following.
The obvious thing to do at this point is to check what the left expression does to forms of type $\phi = f(x,t) \, dx^I$ (1) and $\phi = f(x,t) \, dx^I \wedge dt$ (2). After some computations, one sees that $K$ must satisfy (I)
\begin{equation}
f(x,t) \, dx^I = K\left(\sum_{i\notin I} \frac{\partial f}{\partial x^i} dx^i \wedge dx^I + \frac{\partial f}{\partial t} dt \wedge dx^I\right) \pm d\left(K\left(f(x,t) \, dx^I \right) \right)
\end{equation}
and (II)
\begin{align}
\left[f(x,t) - e(t)\cdot \left(\int_{\mathbb R} f(x,t) \, dt\right)\right] dx^I \wedge dt &= K\left(\sum_{i\notin I} \frac{\partial f}{\partial x^i} dx^i \wedge dx^I\wedge dt\right) \\ & \quad  \pm d\left(K\left(f(x,t) \, dx^I \wedge dt\right) \right)
\end{align}
We see that the main difference between the first and second input for $K$ is that in the second equation, only forms containing $dt$ appear. And in the first equation, we can get back $f(x,t)$ by integrating $\partial f / \partial t$ and sending everything else to $0$. Well, let's try that and see what comes out! So we define 
$$K(f(x,t) \, dx^I ) = 0, \quad K(f(x,t) \, dx^I \wedge dt) = \left(\int_{-\infty}^t f(x,t) \, dt\right) dx^I$$
In the first line (I), we clearly get back $f(x,t) \, dx^I$ which is what we wanted. What about the second one? Again, after some computations
$$K\left(d\phi \right) - dK(\phi) = \phi$$
So we missed a term in (II)... How can we get this summand? What if we just add it in the operator $K$? That would not work out, since we have to differentiate in (II). So a better bet would be to try to exchange a primitive $E(t)$ of $e(t)$ for $e(t)$. We find that we indeed get the result this with
$$K( f(x,t) \, dx^I ) = 0,\quad K(f(x,t) \, dx^I \wedge dt) = \left(\int_{-\infty}^t f(x,t) \, dt \right) \, dx^I - E(t) \cdot \left(\int_{\mathbb R} f(x,t) \, dt \right) \, dx^I $$
Hence this is our operator. 
