Proving $\bigwedge^k(U\cap V) = \left(\bigwedge^kU\right) \bigcap \left(\bigwedge^k V\right)$ via the universal mapping property Let $V$ be a (finite dimensional) vector space, its exterior algebra of order $k$ is the vector space $\bigwedge^k V$ consisting of the formal sums of terms of the form $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where each $v_i \in V$ (with a few additional properties regarding $\wedge$). There are quite a few ways to define $\bigwedge^k V$ rigorously, one of them is via the universal mapping property:

There exists a vector space $L$ and an alternating multilinear map $M\colon V^k \to L$ with the universal mapping property in the sense that for any alternating multilinear map $M'\colon V^k \to L'$, there exists a unique linear map $T\colon L\to L'$ such that $M' = T\circ M$. This space $L$ is unique up to isomorphism and we define $\bigwedge^k V := L$ and write $v_1 \wedge v_2 \wedge \dots \wedge v_k:= M(v_1,v_2,\dots,v_k)$.

From the above construction, all the usual properties of $\bigwedge^k V$ follow, e.g. if $\{ e_1, \dots , e_n \}$ is a basis for $V$, then $\{ e_{i_1} \wedge \dots \wedge e_{i_k} \}_{i_1<\dots<i_k }$ is a basis for $\bigwedge^k V$. If $U$ is a subspace of $V$, then $\bigwedge^k U$ can be canonically identified with a subspace of $\bigwedge^k V$ via the universal property (the inclusion map $i\colon U^k\to V^k$ composes with $M$ is an alternating multilinear map from $U^k$ into $\bigwedge^k V$).
Now, let $W$ be a vector space and $U,V$ be subspaces of $W$. I believe it is true that $\bigwedge^k(U\cap V) = \left(\bigwedge^kU\right) \bigcap \left(\bigwedge^k V\right)$, which should be straight forward to prove  by fixing a common basis for $U,V$ in $W$. However, I want to know if it can be demonstrated from the perspective of category theory using the universal mapping property. It seems that this would require some characterization of $U\cap V$ in terms of morphisms but my working knowledge of techniques  from category theory has mostly faded away at this point (not that there was much of it from the beginning). Any help is highly appreciated, especially if it's accompanied by a diagram :-)
 A: Let me give some evidence that you should not expect there to be a purely "formal" proof using the universal property.  Namely, an analogous result is not true in a slightly more general setting.
Specifically, instead of vector spaces over a field, let us consider $\mathbb{Z}$-modules.  Let $W=\mathbb{Z}^2$ with standard basis $\{e_0,e_1\}$, let $U$ be the submodule generated by $2e_0$ and $e_1$, and let $V$ be the submodule generated by $e_0$ and $2e_1$.  Then $\bigwedge^2W$ is freely generated by $e_0\wedge e_1$, and the inclusion maps $U,V\to W$ induce injections on the second exterior powers that identify $\bigwedge^2 U$ with the submodule generated by $2e_0\wedge e_1$ and $\bigwedge^2 V$ with the submodule generated by $e_0\wedge 2e_1$.  But $e_0\wedge 2e_1=2e_0\wedge e_1$ in $\bigwedge^2 W$ so these are the same submodule.  On the other hand, $U\cap V$ is generated by $2e_0$ and $2e_1$, so $\bigwedge^2(U\cap V)$ is the submodule of $\bigwedge^2 W$ generated by $2e_0\wedge 2e_1=4e_0\wedge e_1$.  So in this case, $\bigwedge^2(U\cap V)$ is strictly smaller than $\bigwedge^2U\cap \bigwedge^2 V$ as submodules of $\bigwedge^2 W$.
So, any proof of this property is going to need to use something that is true about vector spaces but not true about $\mathbb{Z}$-modules (or even just finitely generated free $\mathbb{Z}$-modules).  Basically, this means you need to use bases, or at least semisimplicity (i.e., you need to use the fact that $W$ can be decomposed as a direct sum $W_0\oplus W_1\oplus W_2\oplus W_3$ where $U=W_0\oplus W_1$ and $V=W_0\oplus W_2$).
A: The important result is
$$\wedge^{\cdot}(V\oplus W)\simeq \wedge^{\cdot}(V)\otimes \wedge^{\cdot}(W)$$
valid for modules over every commutative ring.
However, it is possible to have    modules $U\subset V$, while $\wedge^2(U) \ne 0$, $\wedge^2(V)= 0$. Indeed, consider $U = (\frac{1}{n}\mathbb{Z}/\mathbb{Z})^{\oplus 2} \subset (\mathbb{Q}/\mathbb{Z})^{\oplus 2} = V$.
