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Assume that the $C$ in $\text{Hom}_C(x,y)$ can always be inferred from $x,y$ so that we can change our notation to $\text{H}(x,y) := \text{Hom}_C(x,y)$

Then the Yoneda lemma "looks at a single step of iteration" of $\text{H}$:

$\text{H}(\text{H}(\cdot, x), A) \simeq A(x)$

is the basic statement of Yoneda. Do isomorphisms of higher iterates of $\text{H}$ now become trivial under Yoneda's lemma, or is there a finite iteration of $\text{H}$ in either argument that yields an interesting isomorphism?

For example, what can we say about:

$$ \text{H}(\text{H}(\text{H}(\cdot, x), B), A) $$ is it just that it's isomorphic to $H(B(x), A)$ because of yoneda? Or is there something else we can say?


Following along the lines of proof of Yoneda in Categories & Sheaves, we have a sequence of maps:

$$ \text{H}(\text{H}(\text{H}(\cdot, x), B), A) \to \text{H}(H(H(x,x), B(x)), A(x)) $$

but in order to use the identity trick we $B(x) = \text{Hom}(x,x)$ as well. So what about $B = \text{Hom}(x, \cdot)$. Then we have that we're looking at the finite iterate:

$$ \text{H}(\text{H}(\text{H}(\cdot, x), \text{H}(x, \cdot)), A) \simeq A \circ \text{H}(x,x) $$

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  • $\begingroup$ $\mathsf{Hom}(-,X)$ and $\mathsf{Hom}(X,-)$ are objects of different categories, so you can't talk about the set of arrows between them. $\endgroup$ Nov 24, 2018 at 7:14

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You have to be careful to distinguish both $Hom$ functors: the inner one in the Yoneda lemma is just in your category $C$, while the outer one is in the functor category $Set^{C^{op}}$. Adding another $Hom$ means that you would have to take morphisms between natural transformations (called modifications), which are trivial in $Cat$ (see this Mathoverflow question), so "Yoneda-like" expressions with higher iterates of $Hom$ are always trivial, regardless of Yoneda's lemma.

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  • $\begingroup$ I think it has to do with the fac that the higher isomorphisms are just special cases of the original Yoneda isomorphism since I'm just varying the arguments. $\endgroup$ Nov 24, 2018 at 5:48

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