exchange the exterior derivative with integral I was trying to prove the homotopy invariance of the deRham cohomology ,which involove the following relationship in the construction of the homotopy operator,which I don't know how to prove.
Let $M$ be smooth manifold,consider $M\times I$ with $I = [0,1]$, assume $\omega \in \Omega ^k (M\times I)$.Let $i_t:M \to M\times I$ such that $i_t(p) = (p,t)$
Prove that
$$d\int i^*_t (\omega)dt = \int di^*_t(\omega)dt$$
I try to prove it as follows:
For simplicity it's enough to prove the case for differential 1- form,
Since  under the local coordinate ,we have $\omega = \omega_idx^i + \omega_{n+1}dt$ so under the pull back we have $i^*_t(\omega) = \omega_i\circ i_t dx^i$, so $$d \int i^*_t(\omega) dt = d((\int\omega_i\circ i_tdt) dx^i) = (\frac{\partial}{\partial x^j}\int \omega_i\circ i_t)dx^j\wedge dx^i$$
Now we exchange integral and differential in the last terms ,which gives the result.
Is my proof correct?
 A: The expression $i_t^*(\omega)$ does not make sense since $i_t^* : \Omega^k(M \times [0,1]) \rightarrow \Omega^k(M)$ and $\omega \in \Omega^k(M)$.
Nevertheless if $\omega \in \Omega^k(M \times [0,1])$, the integral of $\omega$ (no need to put the $dt$ because it is included in $\omega$) is
$$
\int_0^1 : \Omega^k(M \times [0,1]) \rightarrow \Omega^k(M)
$$
As you said, if $\omega$ is a $1$-form then locally it has the form $$
\omega_1(x,t)dx^1 + \dots + \omega_n(x,t)dx^n+ \omega_{n+1}(t,x)dt
$$
where $n$ is the dimension of $M$ and the $\omega_i$'s are just (smooth) functions defined locally.
The right-hand side of $i_t(\omega) = \omega_i \circ i_t dx^i$ does not make sense (here it is assumed that $\omega$ is a differential form on $M \times [0,1]$ so $i_t(\omega)$ makes sense). The symbol $\circ$ is traditionally reserved to composition of functions.
The form $i_t^*(\omega) \in \Omega^1(M)$ is the "slice at time $t$" of the form defined on $M \times [0,1]$ so it just expresses as $\sum_{i=1}^n \omega_i(x,t)dx_i$ where $t$ is fixed, in particular it does not contain a term with $dt$. If $t$ varies, the form $i_t^*(\omega)$ (also usually written $\omega_t$) will be a $1$-form on $M$ varying smoothly with respect to the variable $t$.
Now, the integral can be described as follows: the form $\omega \in \Omega^1(M \times [0,1])$ can be decomposed as $\omega= \alpha + f dt$ where $\alpha \in \Omega^1(M)$ and $f \in C^\infty(M \times [0,1])$
$$
\int_0^1 \omega = \int_0^1 \alpha(x,t) + f(x,t) dt = \int_0^1 f(x,t) dt
$$
Using that the differential $d = \sum_i \frac{\partial}{\partial x^i}$ in local coordinate you should be able to show that
$$
d \left( \int_0^1 \omega\right) + \int_0^1 d \omega = i^*_1(\omega) - i^*_0(\omega).
$$
