Let G be a lie group and let A be a G-differential complex.
Consider the tensor product $A \otimes \wedge \mathfrak{g}^* $. We write an element $\alpha \in A \otimes \wedge \mathfrak{g}^* $ as $\alpha = \sum_k \alpha_{[k]}$ with $\alpha_{[k]} \in A \otimes \wedge^k \mathfrak{g}^*.$ we denote by $r : A \otimes \wedge \mathfrak{g}^* \rightarrow A$ the projection of an element $\alpha \in A \otimes \wedge \mathfrak{g}^*$ on its component $\alpha_{[0]}$ of exterior degree 0.
Please help me to understand the proof of following lemma:
Let A be a G-differential complex. The map r: $A \otimes \wedge \mathfrak{g}^* \rightarrow A$ induced an isomorphism from ${(A \otimes \wedge \mathfrak{g}^* )}_{horG}$ to A.
To the prove the injectivity of this map the authors write:
For $ X \in \mathfrak{g}$, we denote by $\iota_t(X)$ the tensor product contraction on $A \otimes \wedge \mathfrak{g}^*$. If $\alpha \in {(A \otimes \wedge \mathfrak{g}^* )}_{horG}$ is such that $\alpha_{[0]}= 0$, it is easy to see by induction on the exterior degree that $\alpha =0.$
In this proof I'm not sure how the operator $\iota_t(X)$ acts on the elements of the tensor product : we have a contraction on $A$ and a contraction on $\wedge \mathfrak{g}^*$, what is the total contraction on $A \otimes\wedge \mathfrak{g}^* $ (it's not described before this lemma ). And why does $\alpha_{[0]}= 0$ implies that $\alpha=0$ ?