So apparently I'm doing something wrong, but can't figure out what. I've read a proof of the Yoneda lemma and understand it from Riehl's book, but to try it out, I tried an example.
My input category: C has objects X and Y only, with just the arrows, $id_X : X →X$ $g : X →X$ $f : X→Y$ And of course $Y$'s $id$ arrow. I choose $g^2=id_X$, and with only $f:X→Y$, $f$ composed with $g$ must be $f$.
I pick a functor $A:C→Set$, such that $AX =$ {0,1}, and $Ag = id_{AX}$, and as far as I can tell everything up to now is legal.
With $g$ defined the way it is, then $g^{*}: C(X,X) → C(X,X)$ has, $g^{*}(g) = g^2 = id_X$ $g^{*}(id_X) = g$, So $g^*$ swaps the elements of $C(X,X)$. But as I've defined $AX$ and $Ag$, I can't see a natural transformation $α:C(-, X) →A$ that seems to commute the square for $g$, when according to Yoneda there should exist two natural transformations BC $AX$ has two elements!
Further, when I formed the natural square for $f:X→Y$, I get four functions mapping $C(X,X) →AX$ and the $Y$ component $α_Y$ can be whatever that fits, so I figured doing squares for $id_X$ and $g$ would restrict to two for $α_X$, which is when I found the first issue I mentioned.
Can someone please spell this out, I feel like it really shouldn't be this complicated... Edit: Fixed it from 'there is no" to "I can see".
Riehl (like other sources) state Yoneda using $C(X, -)$ rather than $C(-,X)$ (of course, there's a dual version with $C(-, X)$ - see Exercise 2.2.i) - I'm not sure if this is a typo or if you really want the dual version. Assuming it's a typo:
What I think you're concerned about is the following diagram: $$ \begin{array}{ccc} C(X,X) & \stackrel{\alpha_X}\to & A(X)\\ \downarrow C(X,g) & & \downarrow A(g)=\operatorname{id} \\ C(X,X) & \stackrel{\alpha_X} \to & A(X) \end{array} $$ Is that right? You say there aren't any natural transformations $\alpha$ making it commute, but there are two. One $\alpha_X$ sends both $\operatorname{id}_X$ and $g$ to 0, the other sends them both to 1. This is what Yoneda says should happen - the natural transformations are in bijection with $A(X)=\{0,1\}$.