# If the Yoneda lemma essentially says that $\text{Hom}(\text{Hom}(\cdot, x), A) \simeq A(x)$, then what about higher iterates of $\text{Hom}$?

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:

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

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

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.