Dualising the Yoneda lemma.

Question

Let $\mathcal{C}$ be a small category, and $A, B \in \mathcal{C}_0$ objects of $\mathcal{C}$. Suppose that for every $X ∈ \mathcal{C}_0$ we have bijections $f_X : \text{Hom}_{\mathcal{C}}(A, X) \tilde\to \text{Hom}_{\mathcal{C}}(B, X)$ such that postcomposition with every $g : X \to X'$, 'commutes with the $f_X$': $$ (g \circ -) \circ f_X = f_{X'} \circ (g \circ -). $$ That is, for each $h \in \text{Hom}_{\mathcal{C}}(A, X)$, we have $$g \circ f_X(h) = (g \circ -) \circ f_X(h) = f_{X'} \circ (g\circ -)(h) = f_{X'}(g \circ h). $$ Everything typechecks, so it must be right ;).

All this is to say that there exists a natural transformation $\mu \in [\mathcal{C}, \textbf{Set}]_1$ between $\text{Hom}_{\mathcal{C}}(A, -)$ and $\text{Hom}_{\mathcal{C}}(B, -)$, namely

$$\mu = \left(f_X : \text{Hom}_{\mathcal{C}}(A, X) \tilde\to \text{Hom}_{\mathcal{C}}(B, X) \right)_{X \in \mathcal{C}_0} : \text{Hom}_{\mathcal{C}}(A, -) \, \tilde\Longrightarrow\, \text{Hom}_{\mathcal{C}}(B, -). $$ As it is 'point-wise' an isomorphism (of sets), it is a natural isomorphism.

I always get a bit confused around opposite categories, but does this allow me to immediately write down the natural isomorphism

$$\mu^{\text{op}} : \text{Hom}_{\mathcal{C}^{\text{op}}}(B, -) = \text{Hom}_{\mathcal{C}}(-, B) \, \tilde\Longrightarrow \, \text{Hom}_{\mathcal{C}}(-, A) = \text{Hom}_{\mathcal{C}^{\text{op}}}(A, -)\,? $$

If so, is this an isomorphism in the functor category $[\mathcal{C}, \textbf{Set}]^\text{op} $? And is that $[\mathcal{C}^\text{op}, \textbf{Set}]$?

What I ultimately want is to use the fully faithfulness of the Yoneda embedding $y : \mathcal{C} \to [\mathcal{C}^\text{op}, \textbf{Set}]$ to prove that $A \cong B$ in $\mathcal{C}$. However, what I have seems to be an isomorphism between something to the tune of $y^\text{op}(A)$ and $y^\text{op}(B)$ or $y(A)^\text{op}$ and $y(B)^\text{op}$, but I'm not sure how to interpret that or if that's in fact correct.

I feel like I'm almost there, but the dualising keeps confusing me and I can't find the mistake.

EDIT: perhaps the question just boils down to whether $\text{Hom}_{\mathcal{C}}(B, -)^{\text{op}} = \text{Hom}_{\mathcal{C}^{\text{op}}}(B, -)$ as functors.

Answer 1

Note that the $\mu^{op} : Hom_{\mathcal{C}}(B,-) \rightarrow Hom_{\mathcal{C}}(A,-)$ need not be a natural transformation. This is just an arrow in the dual category of $[\mathcal{C}, Set]$ corresponding to the arrow $\mu$. You can only say that this is an isomorphism in the dual category.

The categories $[\mathcal{C}, Set]^{op}$ and $[\mathcal{C}^{op}, Set]$ are completely different.

The functor $Hom_{\mathcal{C}}(B,-)^{op} : \mathcal{C}^{op} \rightarrow Set^{op}$ (this is what the opposite of a functor means. Not to be confused with an object of the dual of $[\mathcal{C}, Set]$).

On the other hand, $Hom_{\mathcal{C}^{op}}(B,-) : \mathcal{C}^{op} \rightarrow Set$ is just the functor $Hom_{\mathcal{C}}(-,B)$.

They are different functors.

If you are trying to obtain a "contravariant version" of the Yoneda lemma, you may proceed as follows :

You already have the covariant version.

Now, take an object $B \in \mathcal{C}$ and consider a presheaf $X : \mathcal{C}^{op} \rightarrow Set$. Consider the representable presheaf $Hom_{\mathcal{C}}(-,B)$ and treat it as a covariant functor $Hom_{\mathcal{C}^{op}}(B,-)$. Then, using the covariant version of the Yoneda lemma tells you that $$Nat(Hom_{\mathcal{C}}(-,B), X) = Nat(Hom_{\mathcal{C}^{op}}(B,-), X) \cong XB$$. This is the contravariant version of the Yoneda lemma.

In particular, it says that the Yoneda functor $y : \mathcal{C} \rightarrow [\mathcal{C}^{op}, Set]$ is fully faithful, so that $y$ reflects isomorphisms (as you wanted).

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A word of caution : There is something called a co-Yoneda lemma (and deals with tensor products of set-valued functors), which is what actually dualizes the Yoneda lemma.

Answer 2

@Category_Theorist: Thanks, that is clarifying! On the one hand, it allows me to say that, by the Yoneda lemma (using the usual form of the Yoneda lemma):

\begin{align*} \text{Hom}_{[\mathcal{C}^\text{op}, \textbf{Set}]}\left(\text{Hom}_{\mathcal{C}}(-, A),\text{Hom}_{\mathcal{C}}(-, B)\right) &\cong \text{Hom}_{\mathcal{C}}(-, B)(A) = \text{Hom}_{\mathcal{C}}(A, B) \end{align*}

(I use the 'Hom'-notation because 'Nat' could suppress 'op's where they should be.)

while on the other hand (just switching $\mathcal{C}$ and $\mathcal{C}^\text{op}$ everywhere):

\begin{align*} \text{Hom}_{[\mathcal{C}, \textbf{Set}]}\left(\text{Hom}_{\mathcal{C}}(A, -),\text{Hom}_{\mathcal{C}}(B, -)\right) &= \text{Hom}_{[\mathcal{C}, \textbf{Set}]}\left(\text{Hom}_{\mathcal{C}^\text{op}}(-,A),\text{Hom}_{\mathcal{C}^\text{op}}(-, B)\right) \cong \text{Hom}_{\mathcal{C}^\text{op}}(-, B))(A) \\ &= \text{Hom}_{\mathcal{C}^\text{op}}(A, B)) = \text{Hom}_{\mathcal{C}}(B, A). \end{align*}

It seems that $\mu$, which is an isomorphism in $\text{Hom}_{[\mathcal{C}, \textbf{Set}]}\left(\text{Hom}_{\mathcal{C}}(A, -),\text{Hom}_{\mathcal{C}}(B, -)\right)$, is now in bijective correspondence to a morphism $B\to A$. However, to show that this is in fact an isomorphism again, I can no longer use the literal Yoneda functor, i.e., $y : \mathcal{C} \to [\mathcal{C}^\text{op},\textbf{Set}]$, because the 'op's are the wrong way around. The most logical thing to me seems to show that there is just as well a functor, $\tilde{y}$, say, as follows: $\tilde{y} : \mathcal{C}^\text{op} \to [\mathcal{C}, \textbf{Set}]$, which also reflects isomorphisms! But that would be an exercise in and of itself. Or would it...?