# Yoneda Lemma (again)

Let $C^{\wedge} = \text{Fct}(C^{\text{op}}, \text{Set})$. Let $\text{h}_C : C \to C^{\wedge}, \ X \mapsto \text{Hom}_C(\cdot, X)$. Then the Yoneda lemma is:

For $A \in C^{\wedge}$ and $X \in C$, $\text{Hom}_{C^{\wedge}}(\text{h}_C(X), A) \simeq A(X)$.

We have, for $\phi \in \text{Hom}_{C^{\wedge}}(\text{h}_C(X), A)$,

$$\require{AMScd} \begin{CD} \text{h}_C(Y, X) @>{\text{h}_C(f, X)}>> \text{h}_C(Z, X)\\ @V{\phi_Y}VV @V\phi_ZVV \\ A(Y) @>{A(f)}>> A(Z) \end{CD} \tag{1}$$

commutes for every $f : Y \to Z$ in $C^{\text{op}}$. In particular for $Y, Z = X$, associate to $\phi$, $\phi_X(\text{id}_X) \in A(X)$. Thus our map $\varphi : \text{Hom}_{C^{\wedge}}(\text{h}_C(X), A) \to A(X)$ is given by $\varphi : \phi \mapsto \phi_X(\text{id}_X)$.

On the other side, given $s \in A(X)$ we want to associate it to a natural map $\psi(s) \in \text{Hom}_{C^{\wedge}}(\text{h}_C(X), A)$. That is we want a family of maps $\psi(s)_Y : \text{Hom}_C(Y, X) \to A(Y)$. By the functoriality of $A$, $A: \text{Hom}_C(Y,X) \to \text{Hom}_{\text{Set}}(A(X), A(Y))$, and composition with $s$ gives a map into $A(Y)$. Thus $\varphi : s \to A(\cdot)(s)$ hopefully is the inverse map to $\varphi$.

$$\psi \circ \varphi(\phi) = \psi(\phi_X(\text{id}_X)) = A(\cdot)(\phi_X (\text{id}_X))$$

Given $f \in \text{Hom}_C(Y,X)$ we have $(A(f) \circ \phi_X)(\text{id}_X) = ?$

I'm stuck.

Thanks @Sky.

Use this diagram: $$\require{AMScd} \begin{CD} \text{h}_C(X, X) @>{\text{h}_C(f, X)}>> \text{h}_C(Y, X)\\ @V{\phi_X}VV @V\phi_YVV \\ A(X) @>{A(f)}>> A(Y) \end{CD} \tag{2}$$

So that $(A(f) \circ \phi_X)(\text{id}_X) = (\phi_Y \circ \text{h}_C(f, X))(\text{id}_X) = \phi_Y \circ f = \text{h}_C(f, X) \circ \phi_Y$

Still stuck.

it is just a natural transformation.

• @EyesOnBud yeah – Jian Aug 18 '17 at 6:44
• I see what you're saying use $\text{Hom}_C(X,X) \to \dots$ – Shine On You Crazy Diamond Aug 18 '17 at 6:46
• @EyesOnBud What is your meaning?Sorry,I do not understand. – Jian Aug 18 '17 at 6:49
• see my edits of OP above. I'm still not sure... – Shine On You Crazy Diamond Aug 18 '17 at 6:53
• @EyesOnBud I see.Because $Hom_C(f,X)(Id_X)=f$. – Jian Aug 18 '17 at 6:56

$$\varphi(\phi) \equiv \phi_X(\text{id}_X) \\ \psi(s)_Y \equiv A(\cdot)(s), \ \forall Y \in \text{Ob}(C)$$

Then: $$\psi\circ \varphi(\phi) = \psi(\phi_X(\text{id}_X))_Y = A(\cdot)(\phi_X(\text{id}_X)) = \phi_Y\circ \text{h}_C(\cdot, X)(\text{id}_X) = \phi_Y (\text{id}_X (\cdot)) = \phi_Y(\cdot) = \phi_Y$$ since notationally $\phi_Y \equiv \phi_Y(\cdot)$.

The other direction is $\varphi \circ \psi(s) = \psi(s)_X(\text{id}_X) = (A(\cdot)(s))(\text{id}_X) = A(\text{id}_X (s) = \text{id}_{A(X)} (s) = s$.