A category theoretic claim in Mumford's Lectures on Curves on an Algebraic Surface. Edit: I realized that these functors are actually contravariant so that explains why things weren't coming out completely right...  So my attempt clearly makes no sense. I'll fix the details later when I have time...
On p. 108, Mumford makes the following claim:
Suppose we have a diagram (below) where $i$ is the inclusion, all functors from algebraic schemes over $k$ to $Sets$. Here, $h_G=\operatorname{Hom}(-,G)$. We have the following condition:
(#) For any $\alpha\in B(S)$, if $g:T\rightarrow S$ is given, then there is a subscheme $Y\subseteq S$ such that $g^*(\alpha)\in B(T)$ is in $A(T)$ if and only if $g$ factors through $Y$ i.e. $T\rightarrow Y\rightarrow S$ is $g$.
If the following condition is true, there exists a subscheme $G_0\subseteq G$ such that $A(-)\cong h_{G_0}$ and $\Phi$ is induced form the inclusion $G_0\rightarrow G$ to get $\Phi:h_{G_0}\rightarrow h_{G}$.
My question is to on how to prove this result?
My attempt is as follows. Choose $S=G$. Then the look at $g:T\rightarrow G$. Apply the condition above to find that for any $\alpha\in B(G)$, the fibre $g^*(\alpha)\in B(T)$ is in $A(T)$ if and only if $g$ factors through $Y_a$. But that means $i:A(T)\rightarrow B(T)$ is determined by some collection of $Y_a$ which $g$ must factor through via the commutative diagram
Another idea I had but not tried is to note that $A(T)\rightarrow h_G(T)$ is always an inclusion. So I should probably be looking at $T\rightarrow G$ determined from $A(T)$. Then try to prove the factorization.
$\require{AMScd}$
\begin{CD}
A(T) @>>> B(T)\\
@VVV @VVV\\
A(G) @>>> B(G)
\end{CD}
Take the union $Y=\bigcup Y_\alpha$. Then $g:T\rightarrow Y$ corresponds to $A(T)$ in the following sense: given $\alpha\in A(T)$, I can determine an element $\alpha\in B(G)$ which pulls-back to an element of $A(T)$ via the diagram above if and only if $\alpha$ corresponds to $g:T\rightarrow Y$ which sends $T$ into $Y_\alpha$. So I get that the subset of $Hom(T,Y)$ where $T$ lands in some $Y_\alpha$ corresponds to elements of $A(T)$.
What I am struggling with is getting Hom(-,Y)\cong A(-) in general.
I suspect I'd have to use the Yoneda lemma in some way here for me to get this, but I've yet to come up with the complete proof...

 A: The condition says that the arrow $A\to B$ is representable. As you suspect, you can translate it into category theory via the Yoneda lemma. I have drawn a picture for you.

For each arrow $\alpha:h_S\to B$ there is some monomorphism $h_Y\to h_S$ such that for each $g: h_T\to h_S$ the composite $\alpha g$ factors through $A$ if and only if  the arrow $h_T\to h_Y$ exists. This tells us that the square is a pullback (pullbacks can be tested on representables). Now form two new pullbacks like in the diagram below.

The outer rectangle is also a pullback by the pullback lemma. The lower map is $\psi \phi =i$. But if you pull a monomorphism back against itself, then the resulting pullback cone is $A$ with two identities into $A$. So we can arrange it such that the rectangle above looks like this.

From that diagram you can easily read of that both $A\to h_G$ and $h_{G_0}\to h_G$ are two subobjects which factor through each other. From this it follows that they must be isomorphic as subobjects of $h_G$.
