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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?

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Horrible, horrible issues would arise. It is very easy to find non-natural bijections, so I'm not sure what you mean. –  Qiaochu Yuan Aug 16 '10 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. –  G. Rodrigues Aug 16 '10 at 19:32
More generally, one could ask: Why is a natural isomorphism between two functors actually required to be natural? –  Rasmus Aug 16 '10 at 20:37

4 Answers 4

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!

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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.

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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.

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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/.

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