Intuition for the chain homotopy of Poincaré lemma. In the proof of Poincaré lemma, Bredon essentially constructs a chain homotopy between the identity and the null map. Namely, for an open convex set $U \subset \mathbb{R}^{n+1}$ containing the origin,  he builds
$$\phi: \Omega^{p+1} (U) \to \Omega^{p}(U) $$
as follows. For $\omega=fdx_{j_0} \wedge \cdots dx_{j_p}$, define
$$\phi(\omega)=\big(\int_0^1t^pf(tx)\big)\eta ,$$
where $\eta=\sum_{i=0} ^p (-1)^i x_{j_i}dx_{j_0} \wedge \cdots \wedge \widehat{dx_{j_i}} \wedge \cdots \wedge dx_{j_p}.$ Then extend linearly.
Now to check that it is a chain homotopy is just computation. What I don't get is how to come up with that idea.
Is there some intuition/motivation for considering this chain homotopy?
 A: Expanding on the comments: This is a special case of the chain homotopy associated with a homotopy $M \times [0,1] \to N$.
With $\pi \colon M \times [0,1] \to M$ being the projection, we can let $\phi = \pi_* \circ h^*$
$$ \Omega^{p+1}(N) \xrightarrow{h^*} \Omega^{p+1}(M \times [0,1]) \xrightarrow{\pi_*} \Omega^p(M) $$
where $\pi_*$ is integration along the fiber. Then
$$ d\circ \phi + \phi \circ d = h_1^* - h_0^* \colon \Omega^p(N) \to \Omega^p(M)$$
so that $\phi$ is a chain homotopy between the chain maps $h_0^*$ and $h_1^*$.
In your case we have a homotopy $h \colon U \times [0,1] \to U, h(x,t) = tx$ from the constant map to the identity map (which only exists because $U$ is star-convex around $0$). Then we have $h^*(dx_j) = t \,dx_j + x_j \,dt$, so you can work out that, for $\omega=f \,dx_{j_0} \wedge \cdots dx_{j_p}$,
$$h^*(\omega)(x,t) = t^{p+1} f(tx) dx_{j_0} \wedge \cdots \wedge dx_{j_p} + t^p f(tx) \,dt \wedge \eta,$$
where $\eta$ is as you defined. Then integrate:
$$ \phi(\omega)(x) = \pi_*(h^*(\omega))(x) = \int_0^1 t^p f(tx) \,dt \cdot \eta.$$
Now $\phi$ is a chain homotopy between $h_0^* = 0$ and $h_1^* = \operatorname{id}$ and you're done.
