This is a simple algebraic question I feel I should be obvious, but maybe isn't.
Let $d'\colon V \twoheadrightarrow W$ be a surjective linear map of finite-dimensional vector spaces over a field of characteristic 0.
Write $\Lambda V$ for the exterior algebra on $V$ and define an antiderivation on the $\Lambda V$-module $\Lambda V \otimes (k \oplus W)$ by setting
$$d(k \otimes W) = d(k \otimes k) = 0,$$ $$d(v \otimes 1) = 1 \otimes d'v$$ and extending $d$ uniquely to a $k$-antiderivation.
Let $\hat V$ be the kernel of the restriction of $d$ to $V \otimes k$.
I feel strongly that the kernel $\ker d$ should meet the factor $\Lambda V \otimes k$ in the exterior subalgebra $\Lambda \hat V \otimes k$, but don't have a concrete reason why. It's clear $\Lambda\hat V \otimes k \subseteq \ker d$ but the converse is less clear.
I think a proof by contradiction might start with letting $n$ be minimal such that the kernel of $d$ on the subalgebra $\Lambda^n V \otimes k$ spanned by products of $n$ elements of $V$ is not contained in $\Lambda^n \hat V \otimes k$, finding a sum with the the minimum possible number of terms witnessing this failure ... and then somehow contradicting minimality. I'm not sure how one would proceed from this point.
1) Is my feeling even right?
2) Where can I find a proof, if it's correct?
3) What's a counterexample, if it's wrong?
In case it's of interest, my motivation for asking this question is that it's always the case that if $G$ is a compact connected Lie group and $K$ a closed subgroup, then in the Serre spectral sequence of the Borel fibration $G \to G/K \to BK$, the leftmost column of the $E_\infty$ page is an exterior algebra on a subspace of the space $PG \leq H^*(G)$ of primitives of $G$ (where $H^*(G) \cong \Lambda[PG]$. This means that the differentials $d_r$ in the spectral sequence act as in the question. But I believe this is usually proven by other means and wondered if there were a purely algebraic proof of this fact with no topology, Lie algebras, or indeed Hopf algebras involved.