I'm been trying to solve the following problem form Warner ex 2.15:
# 15: Let $\eta \in V$. Prove that the composition
$$ \Lambda^p(V) \xrightarrow{\eta \wedge}{} \Lambda^{p+1}(V)\xrightarrow{\eta \wedge}{} \Lambda^{p+2}(V) $$ of left exterior multiplication by $\eta$ with itself is an exact sequence.
Solution: First, let $\omega \in \Lambda^p(V)$, we observe that $$ [(\eta \wedge ) \circ (\eta \wedge)](\omega)=(\eta \wedge \eta)\wedge \omega=0 \Rightarrow \ker (\eta \wedge) \supset im (\eta \wedge). $$ Second, the hardest part is to prove that $\ker (\eta \wedge) \subset im (\eta \wedge):$ We must prove that if $(\eta\wedge)(\omega)=0$ for $\omega \in \Lambda^{p+1}(V)$, then there exists $\zeta \in \Lambda^p(V)$ such that $(\eta \wedge)(\zeta)=\omega:$ any clue/hint for this part will be wonderful.
Following your hints. I establish the following
Extending $\eta$ to a basis $\eta_1,...,\eta_n$ for $V^{*}$ with $\eta_1=\eta$ and set $$ \omega=\sum_{J}c_J\eta_{J} $$ where $J$ runs over all strictly ascending multi-indices $1\leq j_1<\cdots <j_p\leq n$. In the sum $$ \eta \wedge \omega=\sum_{J}\eta \wedge \eta_{J} $$ all terms $\eta \wedge \eta_{J}$ with $j_1=1$ vanish since $\eta=\eta_1$. Hence, $$ 0=\eta \wedge \omega=\sum_{j_1 \neq 1} c_J \eta \wedge \eta_J. $$ Since $\{\eta \wedge \eta_J\}$ for $j_1 \neq 1$ is a subset of a basis for $\Lambda^{p+1}(V)$, it is linearly independent, and so all $c_J$ are $0$ if $j_1 \neq 1$. Therefore, $$ \omega= \sum_{j_1=1}c_J\eta_J =\eta \wedge \underbrace{\left( \sum_{j_1=1}c_J \eta_{j_2}\wedge \cdots \eta_{j_p} \right)}_{\zeta}. $$ Thereby, $\ker (\eta \wedge)=im (\eta \wedge)$ i.e., it is an exact sequence.
Thank you so much for your time. I do appreciate your help.