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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 :-)

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  • $\begingroup$ This is curious from the categorical point of view. Let's take the exterior algebra $\bigwedge^{\ast}$ rather than just one $\bigwedge^k$. This furnishes a functor from the category of vector space to the category of graded commutative algebras that is left adjoint to the forgetful functor. Intersections in these categories are pullbacks along the inclusions, so we're asking whether a left adjoint preserves pullbacks, which are limits. But, generically, it's the right adjoints that preserve limits. $\endgroup$
    – Thorgott
    Sep 17, 2021 at 16:27
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    $\begingroup$ @Thorgott it is not uncommon for left adjoints to preserve some limits, and in fact it comes up naturally quite often. For example, a good notion for a morphism between topoi is a left adjoint which preserves finite limits. $\endgroup$
    – Brian Shin
    Sep 17, 2021 at 17:02

2 Answers 2

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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 $V=W_0\oplus W_2$).

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  • $\begingroup$ Very interesting. In any case, is it possible that there is some property intrinsic to the category of (finite dimensional) linear spaces over $\Bbb R$ that would perhaps enable the statement to be proved in a category-theory-ish way? $\endgroup$
    – BigbearZzz
    Sep 17, 2021 at 17:17
  • $\begingroup$ I just saw your edit. That does seem to address my question, thanks. $\endgroup$
    – BigbearZzz
    Sep 17, 2021 at 17:18
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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 $U = (\frac{1}{n}\mathbb{Z}/\mathbb{Z})^{\oplus 2} \subset (\mathbb{Q}/\mathbb{Z})^{\oplus 2} = V$.

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    $\begingroup$ Sorry if this is a stupid question but could you please elaborate a bit on how your answer is related to my question? $\endgroup$
    – BigbearZzz
    Sep 17, 2021 at 18:46
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    $\begingroup$ For $\mathbb{Z}$ modules, rather than vector spaces, you can have $U\subset V$ but $\wedge^2(U) \not \subset \wedge^2(V)$. So the argument for vector spaces has to include a bit more than the universality ... it's what the other answer says... $\endgroup$
    – orangeskid
    Sep 17, 2021 at 18:50
  • $\begingroup$ I see what you meant now, thank you. $\endgroup$
    – BigbearZzz
    Sep 17, 2021 at 18:55

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