# The Poincare Lemma for Compactly Supported Cohomology

I´m reading the proof of The Poincare Lemma for Compactly Supported Cohomology there is a part in the proof that said in the text book Bott and Tu: $d \pi_{\ast} = \pi_{\ast} d$ in other words, $\pi_{\ast}: H^{\ast}_c \rightarrow H^{\ast -1}_c$. To produce a map in the reverse direction, let $e = e(t)$ be a compactly supported 1-form on $\mathbb{R}^1$ with total integral 1 and define: $$e_{\ast}:\Omega^{\ast}_c(M) \rightarrow \Omega^{\ast}_c(M \times \mathbb{R}^1)$$ by $\phi \rightarrow (\pi^{\ast}\phi \wedge e)$.

The map $e_{\ast}$ clearly commutes with d, so it also induces a map in cohomology.

Here the text book continue with more a formation but here I have a question:

Since $d \pi_{\ast} = \pi_{\ast} d$ , i.e, $\pi_{\ast}$ is a chain map, then we have the next diagram:

$$\begin{array} A\Omega^{q}_c(M \times \mathbb{R}^1) & \stackrel{d}{\longrightarrow} & \Omega^{q+1}_c(M \times \mathbb{R}^1) \\ \downarrow{\pi_{\ast}} & & \downarrow{\pi_{\ast}} \\ \Omega^{q-1}_c(M) & \stackrel{d}{\longrightarrow} & \Omega^{q}_c(M) \end{array}$$

The question how can i see that this diagram commutes.

I assume that the map $\pi_*$ is the integration along the ${\mathbb R}$-direction. Then commutation with the exterior differential comes from the fact that you can "differentiate under the integral sign": $$\frac{\partial }{\partial x_i} \int_{a}^{b} f(x_1,...,x_n)dx_1= \int_{a}^{b} \frac{\partial }{\partial x_i} f(x_1,...,x_n)dx_1$$ for $i>1$. In addition, you use the fundamental theorem of calculus to take care of the contribution of $\frac{\partial }{\partial x_1}$ in the exterior differential of $\omega\in \Omega^*$; this way you see that the corresponding summand integrates to zero.
Edit: Here is the FTC part: If $\omega\in \Omega^*(R\times M)$ (I am using the $R$ factor as the 1st coordinate) then $d\omega$ might contain the summand
$$\frac{\partial }{\partial x_1}f_1(x_1,...,x_n) dx_1\wedge...$$
Applying $\pi_*$ to this term and the fact that $f_1$ is compactly supported, we get $$\int_{R} \frac{\partial }{\partial x_1}f_1(x_1,...,x_n) dx_1\wedge... = 0$$ in view of the fundamental theorem of calculus. On the other side, when you are considering $d\pi_*\omega$, since $x_1$-variable is absent, there is no contribution from $dx_1$.
• Excuse another question in this maybe is a stupid question but why $\pi_{ast}: \Omega^{q}_c(M \times \mathbb{R}^1) \rightarrow \Omega^{q-1}_c(M )$ is a chain map? – Knight Nov 6 '13 at 21:12