# Does the forgetful functor from $\mathbf{VEC}_{\mathbf{R}}$ to $\mathbf{SET}$ have right adjoint?

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!

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

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.

• @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. Commented Dec 11, 2022 at 20:45
• @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.
– Alex
Commented Dec 11, 2022 at 21:37
• @MartinBrandenburg Understood, thanks for the advice :)
– Alex
Commented Dec 11, 2022 at 21:38
• @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).) Commented Dec 12, 2022 at 3:07

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.)

• 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 $F(V ⊕ W)$. Commented Dec 11, 2022 at 14:52
• 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. Commented Dec 11, 2022 at 15:40
• 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 $F(V ⊕ W) = F(V) × F(W)$. Commented Dec 11, 2022 at 16:23
• Oh, that's a fair point, I'll switch those. Commented Dec 11, 2022 at 18:16
• @DanielHast Yes, indeed! Thank you very much, this is really a fairly general argument!
– Alex
Commented Dec 11, 2022 at 21:32