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Let $\mathbf{SET}$ be a category of sets, and $\mathbf{VEC}_{K}$ be a category of vector spaces over a field $K$.

In my course on Category Theory we discussed the concept of adjoint functor.

For example, we have constructed a functor $F$ that is left adjoint to the forgetful functor from $\mathbf{VEC}_{K}$ to $\mathbf{SET}$. Indeed, it is enough to take a functor that sends every set $A$ to some vector space with a basis $A$.

But what about the right adjoint functor to this forgetful functor in the case of $K$ equal to the field of real numbers $\mathbf{R}$?

After several unsuccessful attempts to come up with such a functor, it began to seem to me that such a functor does not exist at all.

Is it so? If so, why? If not, then how to build such a functor?

Any hints or advices would really help me, thank you!

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  • $\begingroup$ There's no difference between the cases of generic $K$ and specific $\Bbb R$. $\endgroup$ Commented Dec 11, 2022 at 14:11
  • $\begingroup$ Ok, but even in the case of a real field, unfortunately, I have no ideas. That's why I made a correction. $\endgroup$
    – Alex
    Commented Dec 11, 2022 at 14:16

2 Answers 2

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If the functor forgetful functor $F$ were to admit a right adjoint, then it would preserve initial objects. Therefore, it would need to map the zero vector space to the empty set. This is clearly not the case.

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    $\begingroup$ @Alex This is something you need to remember from this answer - it will help you in many similar situation as well: any left adjoint preserves colimits, so being cocontinuous (=colimit-preserving) is the first thing you need to think of and check when you are interested in the existence of a right adjoint. $\endgroup$ Commented Dec 11, 2022 at 20:45
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    $\begingroup$ @JendrikStelzner Thank you! As far as I understood, the fact that such a functor should preserve the initial objects follows from the fact that it should preserve the colimits, as DanielHast wrote. Since every initial object is a colimit of an empty diagram. $\endgroup$
    – Alex
    Commented Dec 11, 2022 at 21:37
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    $\begingroup$ @MartinBrandenburg Understood, thanks for the advice :) $\endgroup$
    – Alex
    Commented Dec 11, 2022 at 21:38
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    $\begingroup$ @Alex: Yes, preservation of initial objects is a special case of preservation of colimits. However, this special case is more fundamental than the general case of colimits: if $F \colon 𝒜 \to ℬ$ is left adjoint to $G \colon ℬ \to 𝒜$, and $I$ is initial in $𝒜$, then $ℬ(F(I), B) ≅ 𝒜(I, G(B)) ≅ \{*\}$ for every object $B$ of $ℬ$, which means that $F(I)$ is initial in $ℬ$. (For example, in Leinster’s Basic Category Theory, preservation of initial objects is shown in 2.1.15 (page 50), whereas preservation of colimits is only shown in 6.3.1 (page 158).) $\endgroup$ Commented Dec 12, 2022 at 3:07
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If a functor $F$ has a right adjoint, then $F$ preserves all colimits. In particular, $F$ must preserve coproducts. The coproduct in the category of vector spaces is direct sum, while the coproduct in the category of sets is disjoint union.

However, the forgetful functor $F\colon \mathbf{VEC}_K \to \mathbf{SET}$ does not preserve coproducts: For vector spaces $V$ and $W$, we have a diagram $$V \to V \oplus W \leftarrow W,$$ where the morphisms are the natural inclusions $v \mapsto (v, 0)$ and $w \mapsto (0, w)$, satisfying the universal property of coproducts. If we apply $F$ to this diagram, the resulting diagram of sets does not satisfy the universal property of coproducts, since the induced function $$F(V) \sqcup F(W) \to F(V \oplus W) = F(V) \times F(W)$$ is not a bijection (that is, an isomorphism of sets) unless $V$ or $W$ is the zero space.

Thus, this forgetful functor does not have a right adjoint. This illustrates a general technique for proving that adjoints don't exist: find a limit (respectively, colimit) that isn't preserved to show a functor doesn't have a left (respectively, right) adjoint. (As a side note, there is a converse, but it requires an additional smallness hypothesis; this is known as the adjoint functor theorem.)

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  • $\begingroup$ I think the sequence $F(V) ⊔ F(W) \to F(V) × F(W) = F(V ⊕ W)$ would make more sense if you removed the term $F(V) × F(W)$, or swapped the terms $F(V) × F(W)$ and $F(V ⊕ W)$. $\endgroup$ Commented Dec 11, 2022 at 14:52
  • $\begingroup$ I'm not sure I see the issue with how it's written? Anyway, your answer is more straightforward—for some reason I forgot to just look at initial objects. $\endgroup$ Commented Dec 11, 2022 at 15:40
  • $\begingroup$ Preservation of binary coproducts for a functor $F \colon 𝒜 \to ℬ$ means that the natural morphism $F(A) ⨿ F(B) \to F(A ⨿ B)$ is an isomorphism for all $A, B$ (if $𝒜$ and $ℬ$ have binary coproducts). So in our case, I expect to see the natural map $F(V) ⊔ F(W) \to F(V ⊕ W)$. I find it strange to see $F(V) ⊔ F(W) \to F(V) × F(W)$ instead, since in $\mathbf{Set}$, there is no natural map $A ⊔ B \to A × B$. Basically, we don’t have a map $F(V) ⊔ F(W) \to F(V) × F(W)$ until we have first established the map $F(V) ⊔ F(W) \to F(V ⊕ W)$ and the set-theoretic equality $F(V ⊕ W) = F(V) × F(W)$. $\endgroup$ Commented Dec 11, 2022 at 16:23
  • $\begingroup$ Oh, that's a fair point, I'll switch those. $\endgroup$ Commented Dec 11, 2022 at 18:16
  • $\begingroup$ @DanielHast Yes, indeed! Thank you very much, this is really a fairly general argument! $\endgroup$
    – Alex
    Commented Dec 11, 2022 at 21:32

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