I am having trouble proving a result in Mumfords book 'Lectures on Curves on an Algebraic surface. 


It is a statement about the representability of some functor. It is stated on page 108 and says the following. Let $k$ be a field (does anything change if we consider a noetherian scheme instead?) Let $G$ be a scheme over $k$ and $B$ a functor from the category of schemes over $k$ to sets. Let $A$ be a sub functor of $B$. Consider the following commutative diagram of functors 


$$ 
\require{AMScd}
\begin{CD}
A@>>\Phi> h\_G\\
@VVV @| \\
B @<<\Psi<h\_G
\end{CD}
$$
where the first vertical map is the natural inclusion as sub functors and the right vertical map is the identity. 


Now assume that for each scheme $S$ over $k$. And for all $\alpha \in B(S)$, there is a subschema $Y\subset S$ such that all $g\colon T\to S$ 
the pullback $g^\*(\alpha)\in B(T)$ lies in the subset $A(T)$ if and only if $g$ factors through $Y$.


Then he states that $A$ is representable by some subschema $G\_0\subset G$. 


Well I guess the prove works by setting $S=G$ and $G\_0=Y$. Then $\Phi$ factors through $h\_{G\_0}$. But I do not see why this is a surjective mapping? 


Does the proof work differently or is this general statement just wrong?



Is Mumford's statement about the representability of some functor wrong?
The proof works *slightly* differently; the $G\_0$ one needs may be not the one you name but a **retract** of it. In other words, the map you ask about may be nonsurjective but it has a retraction. 


The assumption can be equivalently formulated as follows: for any $\alpha:h\_S\to B$, the pullback of $i:A\hookrightarrow B$ along it is a representable subfunctor $h\_Y\hookrightarrow h\_S$. In particular, as you noted, the pullback of $i$ along $\Psi:h\_G\to B$ is a representable subfunctor $h\_M\hookrightarrow h\_G$. Now let us further pull this subfunctor back along $\Phi:A\to h\_G$. Since $\Psi\circ\Phi=i$, what we obtain is isomorphic to the pullback of $i$ along itself, i. e. to $A$:
$$ 
\require{AMScd}
\begin{CD}
A@<<< h\_M@<<<A\\
@ViVV @VVV @|\\
B @<<\Psi<h\_G@<<\Phi<A
\end{CD},
$$
and the upper horizontal composite is the identity of $A$. Thus although $A$ might be not isomorphic to $h\_M$, it is a retract of it, hence representable itself by some $G\_0$ (in fact by the kernel/image/cokernel of the idempotent $M\to M$ corresponding by Yoneda to the composite $h\_M\to A\to h\_M$).


PS


By the way it follows (if my argument is correct) that it suffices to restrict the above assumption to $\Psi:h\_G\to B$ only rather than arbitrary $\alpha:h\_S\to B$. However if one looks at the place where Mumford verifies this assumption in the situation he needs it, it is maybe even easier to work with arbitrary $S$ rather than use the specific properties of $G$ (the latter is certain Grassmanian in his situation).


