TITLE: Differential Calculus and the De Rham Homotopy Operator
QUESTION [8 upvotes]: Suppose $M$ is a smooth manifold, $\Omega^k(M)$ the vector space (or $C^\infty(M)$-module in case this is better suited) of $k$-differentialforms on $M$ and
$$I: \Omega^k(M\times \mathbb{R}) \to \Omega^{k-1}(M)$$ the De Rham homotopy operator defined by $I(\omega):= \int_0^1i_{\partial_t}\omega(t)dt$. (Where $t$ is assumed to be the global coordinate of $\mathbb{R}$ and $\partial_t$ its appropriate tangent coordinate)
...
I would like to know, how $I$ interacts with the operators of differential calculus:
First we have the well known equation:
$$d\circ I + I \circ d = i_1^* − i_0^*$$
where $i_j^*$ is the pullback defined by the inclusion 
$i_j : M \to M \times \mathbb{R}; x \mapsto (x,j)$. This is the 'interaction' of $I$ with the exterior differential $d$. 
1.) How does $I$ interacts with the interior product 
$i_X: \Omega^k(M \times \mathbb{R}) \to \Omega^{k-1}(M \times \mathbb{R})$ for a vector field $X$ on $M$ (or $M \times \mathbb{R}$) or (via inclusion) with 
$i_X: \Omega^k(M) \to \Omega^{k-1}(M)$ ?
2.) How does $I$ interact with the exterior product?

Edit: If someone knows a different definition of a homotopy operator, with a known behavior 
related to the interior product $i_X$, I would like to know it, too.

REPLY [2 votes]: If $X$ is a vector field on $M$ (time independent) then 
$i_X \circ I = - I \circ i_X$. If $X$ is also time dependent you can play with the cases $\omega$ exact etc.