To prove Poincaré's lemma, we construct $$ \hat{S}_{p}: \Omega^{p}(U \times \mathbb{R}) \rightarrow \Omega^{p-1}(U) $$ Every $\omega \in \Omega ^{p}(U \times \mathbb{R})$ can be written in the form $$ \omega=\sum f_{I}(x, t) d x_{I}+\sum g_{J}(x, t) d t \wedge d x_{J} $$ where $I=\left(i_{1}, \ldots, i_{p}\right)$ and $J=\left(j_{1}, \ldots, j_{p-1}\right)$. We define $$ \hat{S}_{p}(\omega)=\sum\left(\int_{0}^{1} g_{J}(x, t) d t\right) d x_{J} $$ Then we have that \begin{aligned} d \hat{S}_{p}(\omega)+\hat{S}_{p+1} d(\omega) &=\sum_{J, i}\left(\int_{0}^{1} \frac{\partial g_{J}(x, t)}{\partial_{i}} d t\right) d x_{i} \wedge d x_{J} \\ &+ \color{red}{\sum_{I}\left(\int_{0}^{1} \frac{\partial f_{I}(x, t)}{\partial t} d t\right) d x_{I}-\sum_{J, i}\left(\int_{0}^{1} \frac{\partial g_{J}}{\partial x_{i}} d t\right) d x_{i} \wedge d x_{J}} \\ &=\sum\left(\int_{0}^{1} \frac{\partial f_{I}(x, t)}{\partial t} d t\right) d x_{I} \\ &=\sum f_{I}(x, 1) d x_{I}-\sum f_{I}(x, 0) d x_{I} \end{aligned}
I didn't see where the red term comes from ? any help please !