I've been trying to wrap my head around Yoneda's Lemma but it isn't clicking. I've been reading through this post here but I need some help.
I spent a while starring at this commutative diagram on Wikipedia:
Here is how I'm thinking about this and how it goes astray. I think the outer square is showing us the images of $A$ and $X$ under both functors, then we fix any $f : X \rightarrow Y$ and the inside square follows the behavior of $Hom(A,f)$ and $F(f)$, respectively in order to properly establish $\Phi_*$. So clearly $Hom(A,f)$ maps $id_A$ to $f$ (by definition of $Hom(A,-)$). Now we freely pick $u$ (because the theorem says each such choice leads to a unique $\Phi$). Then we know that $F(f)$ acts on $u$ by taking it to $(Ff)u$ and so we define $\Phi_X$ to take $f$ to that to get commutativity. Fine,but how does this extend to arbitrary $X,Y$ where none of them are $A$ if all we know how to do is map $id_A$? In other words, If on the topmost left corner square we replace $Hom(A,A)$ with $Hom(A,Y)$ then how do we know $\Phi_Y$ and $\Phi_X$ do the right thing, or how do we even know how they behave
Here is a diagram where I attempt to look at what happens for arbitrary $X$ and $Y$. I know I need a $g : X \rightarrow Y$ to get a map $Hom(A,g)$ which must take an $f : A \rightarrow X$ to $g \circ f \in Hom(A,Y)$. Then I know the bottom right corner is $\Phi_Y(g \circ f)$ but I don't know where $f$ goes. Taking it from the Wikipedia square (which took $f$ to $(Ff)u$ I now have something like $F[(Ff)u]$ should equal $\Phi_Y(g \circ f)$?