Kernel of the map $\bigoplus_{k\geq 0} \Lambda^k \pi: \bigoplus_{k\geq 0} \Lambda^k V\longrightarrow \bigoplus_{l\geq 0} \Lambda^l(V/W)$? Let $V$ be a vector space over a field $\mathbb K$ and $U\subseteq V$ a subspace. Let $$\pi:V\longrightarrow V/U,$$ the cannonical projection. For every $k\geq 0$ this induces a linear map $$\Lambda^k \pi:\Lambda^k V\longrightarrow \Lambda^k (V/U),$$ where $\Lambda^0 \pi=\pi$. This allows use to define a map $$\bigoplus_{k\geq 0}\Lambda^k \pi: \bigoplus_{k\geq 0} \Lambda^k V\longrightarrow \bigoplus_{k\geq 0} \Lambda^k (V/U).$$ This is the only linear make which make the diagram below to commute for every $k$:

where $\jmath_k$ and $\jmath_k^\prime$ are the inclusions.
What is the kernel of the map $\displaystyle\bigoplus_{k\geq 0} \Lambda^k \pi$?
Well, I know that: $$\mathsf{Ker}(\bigoplus_{k\geq 0} \Lambda^k \pi)=\bigoplus_{k\geq 0} \mathsf{Ker}(\Lambda^k \pi),$$ so the problem boils down to determining $\mathsf{Ker}(\Lambda^k \pi)$.
Thanks. 
 A: Let $T \colon V \rightarrow W$ be a linear map which is onto. The kernel of $\Lambda(T) \colon \Lambda(V) \rightarrow \Lambda(W)$ is the two-sided homogeneous ideal of $\Lambda(V)$ generated by $\ker(T)$. More explicitly, the kernel of $\Lambda^k(T)$ is
$$ U_k = \operatorname{span} \{ v_1 \wedge \dots \wedge v_k \, | \, v_j \in V, \exists i \text{ such that } v_i \in \ker(T) \}. $$
To see why, note first that $U_k \subseteq \ker \Lambda^k(T)$ because on each $v_1 \wedge \dots \wedge v_k$ with $v_i \in \ker(T)$ we have
$$ \Lambda^k(T)(v_1 \wedge \dots \wedge v_k) = Tv_1 \wedge \dots \wedge Tv_i \wedge \dots Tv_k = Tv_1 \wedge \dots \wedge 0 \wedge \dots Tv_k = 0. $$
Hence, $\Lambda^k(T)$ induces a map $\Lambda^k(T) / U_k \colon \Lambda^k(V) / U_k \rightarrow \Lambda^k(W)$. To show that $\ker \Lambda^k(T) = U_k$, it is enough to show that $\Lambda^k(T) / U_k$ is an isomorphism. 
Let's define a map 
$$S \colon \underbrace{W \times \dots \times W}_{k \text{ times}} \rightarrow \Lambda^k(T) / U_k$$ 
as follows. Given $(w_1,\dots,w_k)$ we can choose $v_1, \dots, v_k \in V$ such that $Tv_i = w_i$ (because $T$ is onto). Then define $S(w_1,\dots,w_k) = [v_1 \wedge \dots \wedge v_k]$. To see that this is well-defined, note that if (say) $Tv_1 = T\hat{v}_1 = w_1$ then
$$ [v_1 \wedge \dots \wedge v_k] = [(v_1 - \hat{v}_1 + \hat{v}_1) \wedge \dots \wedge v_k] = [(v_1 - \hat{v}_1) \wedge \dots \wedge v_k] + [\hat{v}_1 \wedge \dots \wedge v_k] \\= [\hat{v}_1 \wedge \dots \wedge v_k]$$
because $v_1 - \hat{v}_1 \in \ker(T)$ so $\hat{v}_1 \wedge \dots \wedge v_k \in U_k$ and similarly for the other arguments. Also, it is clear that $S$ is alternating so by the universal property of the exterior power it induces a map (still denoted by the same name) $S\colon \Lambda^k(W) \rightarrow \Lambda^k(V) / U_k$ which is the inverse map of $\Lambda^k(T) / U_k$.
