1
$\begingroup$

The (extended) Yoneda lemma is about a natural isomorphism of functors. I'm trying to write down these functors, but am getting stuck. The usual Yoneda lemma gives an isomorphism $\mathsf{Nat}(h_A,G)\cong G(A)$. Replacing $G,A$ with hyphens we have $$\mathsf{Nat}(h_{(-)},-)\cong G(-).$$

The functor $\mathsf{Nat}$, I think, is the bifunctor

$$\mathsf{Nat}:(\hat{\mathsf C})^\text{op}\times \hat{\mathsf C}\rightarrow \mathsf{Set}.$$

The functor $h_{(-)}$, I think, should be functor $$y: \mathsf C \rightarrow \hat{\mathsf C}$$ obtained as the transpose of the twisted hom $$\mathsf{Hom}:\mathsf C\times \mathsf C^\text{op}\rightarrow \mathsf{Set}.$$ However, this gives a contradiction since the codomain of $y$ is $\hat{\mathsf{C}}$ while the left part of the domain of $\mathsf{Nat}(h_{(-)},-)$ is $(\hat{\mathsf C})^\text{op}$.

What's the correct definition?

$\endgroup$
1
$\begingroup$

Take the opposite functor of $y$.

The natural isomorphism described is between the (twisted) evaluation $\operatorname{eval}: \mathsf C^\text{op}\times \widehat{\mathsf C}\longrightarrow \mathsf{Set}$ and the composite below

$$\mathsf C^\text{op}\times \widehat{\mathsf C}\overset{(y^\text{op},1)}\longrightarrow (\widehat{\mathsf C})^\text{op}\times \widehat{\mathsf C}\overset{\mathsf{Nat}}\longrightarrow \mathsf{Set}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy