# Adjoint functors requiring a natural bijection

When showing that two functors $F:A\rightarrow B$ and $G:B\rightarrow A$ are adjoint, one defines a natural bijection $\mathrm{Mor}(X,G(Y)) \rightarrow \mathrm{Mor}(F(X),Y)$. What if one do not require the bijection to be natural, what issues would arise?

• Horrible, horrible issues would arise. It is very easy to find non-natural bijections, so I'm not sure what you mean. Commented Aug 16, 2010 at 19:27
• This is more or less asking what if happens we do not require of a map between two groups to preserve the group operations. The answer is: you would end up with a fairly useless concept. Commented Aug 16, 2010 at 19:32
• More generally, one could ask: Why is a natural isomorphism between two functors actually required to be natural? Commented Aug 16, 2010 at 20:37

Consider this example: take $A$ and $B$ to both be the category of non-zero finite dimensional real vector spaces; then all $\mathrm{Mor}(U,V)$ have the same cardinality, making any two endofunctors adjoint in the unnatural sense!

The heteromorphic treatment of adjoints mentioned in Answer 1 without references can be explored further in Ellerman, David 2006. A Theory of Adjoint Functors—with some Thoughts on their Philosophical Significance. In What is Category Theory? Giandomenico Sica ed., Milan: Polimetrica: 127-183, which can be downloaded at: http://www.ellerman.org/a-theory-of-adjoint-functors/ or for a shorter version: http://www.ellerman.org/adjoint-functors-and-heteromorphisms/.

Adjunction of $F,G$ is a bridge between $\mathcal A$ and $\mathcal B$, in most of the examples so called heteromorphisms are definable from objects of $\mathcal A$ to that of $\mathcal B$, and these have to (naturally!) correspond to elements of both homsets $\mathrm{Mor}(FX,Y)$ and $\mathrm{Mor}(X,GY)$. Then $F$ can be obtained by reflections and $G$ by 'coreflections' in this bigger category which disjointly contains $\mathcal A$ and $\mathcal B$ and the heteromorphisms defined by the adjunction. Naturality is crucial.

• Very interesting ! Could you either elaborate of give a link where the notion of the category disjointly enveloping $\mathcal{A}$ and $\mathcal{B}$ is constructed. Is it like a bipartite graph ? Commented Nov 20, 2020 at 15:03
• Yes, similar, but there can be edges on both sides. See my blog profunctors.zellerede.ml Commented Nov 20, 2020 at 16:01

Given a natural transformation $$\varphi_{} \colon \mathbf A(-,G(-)) \to \mathbf B(F(-),-)$$ this is the same as a family natural transformation between the functors $$\varphi_Y \colon \mathbf A(-,G(Y)) \to \mathbf B(F(-),Y)$$ natural in $Y$.

For every $Y \in \mathbf B$ by yoneda lemma we have that, if $\epsilon_Y=\varphi_Y(1_{G(Y)})$, then $$\varphi_Y(f)=\mathbf B(F(f),Y)(\epsilon_Y)=\epsilon_Y \circ F(f)$$ for every $f \in \mathbf A(X,G(Y))$.

The requirement that $\varphi$ is an isomorphism implies that $\varphi_Y$ are all isomorphisms and so $\epsilon_Y$ must be universal (and that's why the adjoint are important, because they make arise universal objects).

If you drop the naturality condition you cannot use yoneda and so you cannot get the universal morphism.

That's why we need naturality. :)

Hope this helps.