How to show this is a homotopy operator between $f_0$ and $f_1$? Given $f_0, f_1: M \rightarrow N$ maps between smooth manifolds. We defined a homotopy operator between them as a linear map $$ Q: \Omega^{k} (N) \rightarrow \Omega^{k-1} (M)$$ such that $$ f_{1}^{*} - f_{0}^{*} = d \circ Q + Q \circ d$$ holds.
Problem: Let $f_0, f_1: M \rightarrow N$ be smooth maps, let $H: I \times M \rightarrow N$ be a homotopy between them, i.e. $H(0,x) = f_0 (x)$ and $H(1,x) = f_1 (x)$. Also let $I_t : M \rightarrow I \times M: x \mapsto (t,x)$. Prove that $$Q := \int_0^{1} I_t^{*} \circ \iota_{\partial_t} \circ H^{*} dt $$ is a homotopy operator between $f_1$ and $f_0$. Here $\iota$ denotes interior multiplication. 
Attempt: I think I have to use the formula $$ \frac{d}{dt} \rho_t^{*} \alpha_t = \rho_t^{*} (L_{v_t} \alpha_t + \frac{d}{dt} \alpha_t)$$ where $\alpha_t$ is the isotopy of the time-dependent vector field $v_t$ and $L$ denotes the Lie derivative.
Given $\alpha $ a $k$-form on $N$, I wanted to calculate $$ Q (d \alpha) + d (Q \alpha) = \int_{0}^1 (I_t^{*} \circ \iota_{\partial_t} \circ H^{*}) (d \alpha) dt + d  \int_0^{1} (I_t^{*} \circ \iota_{\partial_t} \circ H^{*}) \alpha dt $$ I was not sure how to work this out. Any suggestions/advice?
 A: Suppose you didn't know what $Q$ was yet, but you were given the task to write $(f_1^{*} - f_0^{*}) \alpha$ in the form of $(d  \circ  Q + Q  \circ d)\alpha$, for some $Q$.
Because $d  \circ Q + Q  \circ  d$ contains derivatives, you guess that $Q$ must involve some integrals, so we want to rewrite $f_1^{*}  - f_0^{*}$ as an integral.
But we're given a homotopy between $f_0$ and $f_1$, so we can write $f_1^{*} - f_0^{*}$ as an integral like so:
    $$
    (f_1^{*} - f_0^{*})\alpha = \int_{0}^{1}   \frac{d}{dt} (f_t^{*} \alpha) \: dt
,$$
    where $f_t := H  \circ  I_t$, so $f_t^{*} = (H  \circ  I_t)^{*} = I_t ^{*}  \circ H^{*}$.
    Plugging this in, we get
    $$
    \int_{0}^{1}   \frac{d}{dt} (f_t^{*} \alpha) \: dt= \int_{0}^{1} \frac{d}{dt} I_t ^{*}  \circ H^{*} \alpha \: dt
.$$
    The formula you gave tells us how we can calculate the derivative of the time-dependent pullback of a time-dependent differential form. In this case, we have a time-dependent pullback ($I_t^{*}$) of a constant form $(H^{*}\alpha)$. Using the formula, we can move $\frac{d}{dt}$ through $I_t^{*}$ which then changes into a Lie derivative, $\mathcal L_{\partial_t}$.
    $$
\int_{0}^{1} \frac{d}{dt} I_t ^{*}  \circ H^{*} \alpha \: dt= 
    \int_{0}^{1} I_t ^{*}  \circ \mathcal{L}_{\partial_t} (H^{*} \alpha) \: dt
.$$ 
    Now, use Cartan's magic formula: $\mathcal{L}_X = i_X  \circ  d + d  \circ i_X$, and use that $d$ commutes with pulling back. This way the integral becomes a sum of two integrals, and we can move one $d$ upfront, and in the other term, we can move it at the back.
    \begin{align*}
    \int_{0}^{1} I_t ^{*}  \circ \mathcal{L}_{\partial_t} (H^{*} \alpha) \: dt&= 
    \int_{0}^{1} I_t ^{*}  \circ \left( d  \circ i_{\partial t} + i_{\partial_t} d \right)  \circ  H^{*} \alpha \: dt\\
                                                                         &=
    d \int_{0}^{1} I_t ^{*}  \circ  i_{\partial t}    \circ  H^{*} \alpha \: dt+ 
    \int_{0}^{1} I_t ^{*}  \circ  i_{\partial_t}    \circ  H^{*} d\alpha \: dt\\
                                                                         &= (d  \circ  Q + Q  \circ d)\alpha
.\end{align*} 
    This proves that
    $$
Q := \int_{0}^{1}  I_t^{*}  \circ  i_{\partial t}  \circ H^{*} \: dt
$$ 
    is a homotopy operator.
