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Let $V$ be a vector space over a field $k$.

Define $V \otimes - :$ Vect $\to$ Vect the tensor endofunctor on the category of $k$ vector spaces.

Assume that $V \otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F \dashv V\otimes-$.

It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W \cong \oplus_{i \in I}k$ where a basis of $W$ is indexed by $I$.

Show that $F$ is given by tensoring with some vector space.

Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W \otimes V_W$, for some vector space. Maybe they mean something stronger, that $F(W) = W \otimes Z$ for all $W$.

In order to prove something like that I'd need to find a canonical map $\phi: W \times V_W \to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.

I also have this information:

Hom$(F(W), Z) \cong$ Hom$(W, V\otimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.

Any guidance?

Mariah
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    Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true. – Randall Dec 29 '18 at 14:30
  • A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something – Card_Trick Dec 29 '18 at 15:10
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    I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick – Randall Dec 29 '18 at 15:29
  • @Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question. – Mariah Dec 29 '18 at 15:48
  • I didn’t see that $V$ is fd. – Randall Dec 29 '18 at 15:58
  • @Randall I guess we're not assuming it for this part of the question, though. The above should be true in any case, assuming that $ V \otimes -$ preserves limits, and has a left adjoint. – Mariah Dec 29 '18 at 16:00

2 Answers2

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$V \otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism

$$V \otimes (-) \cong \text{Hom}(V^{\ast}, -)$$

and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{\ast} \otimes (-)$.

In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M \otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $\text{Hom}_S(M, S)$ over $R$; see this blog post.

Qiaochu Yuan
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  • More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint. – egreg Dec 29 '18 at 22:45
  • Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V \otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions? – Mariah Dec 29 '18 at 22:51
  • @Mariah: I'm not assuming it. It's a consequence of the assumption that $V \otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $\text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space. – Qiaochu Yuan Dec 29 '18 at 22:51
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    @QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V \otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring? – Mariah Dec 29 '18 at 23:21
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    Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct. – Qiaochu Yuan Dec 30 '18 at 01:23
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To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $\kappa$, then $F(W)$ is the sum of $\kappa$ copies of $F(k)$. This is the same as $W\otimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)\cong \oplus_\kappa F(k)\cong F(k)\otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.

Kevin Carlson
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  • Hey Kevin, thank you! Can you provide a reference to why $\oplus_{\kappa}F(k) \cong W \otimes F(k)$? I guess I'm lacking some knowledge about tensor products.. – Mariah Dec 30 '18 at 22:48
  • @Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases. – Kevin Carlson Dec 30 '18 at 22:54