This is a rather nice question I got from [this user](https://mathoverflow.net/users/3582/) via private communication.


Let $\mathcal{C} = Top$ the category of topological spaces. Let $\mathcal{C}^\prime$ be the category $Funct(\mathcal{C}^{op}, Sets)$. Say $F \in \mathcal{C}^\prime$ is a
sheaf, with the usual patch up condition. Does this imply that F is
representable?


An elementary way of putting this without using the category theory terminology: Say a set $X$ is endowed with the following data:


For every topological space $Y$, a subset $M(Y,X) \subset Maps(Y,X)$, such that,


1. whenever $f: Y \rightarrow X$ in $M(Y,X)$, and $g : Z \rightarrow Y$ continuous,
$f \circ g : Z \rightarrow X$ belongs to $M(Z,X)$.
2. If ${U\_i}$ is an open cover of $Y$, and $f: Y \rightarrow X$ is a set map,
then $f \in M(Y,X)$ if and only if $f|\_{U\_i} \in M (U\_i, X)$ for all $i$.


Now, endow $X$ with the finest topology such that for all $f: Y \rightarrow X \in
M (Y,X)$ [for all topological space $Y$], $f$ becomes continuous. In
other words, declare $U \subset X$ to be open if for all $f \in M(Y,X)$
for all $Y \in ob (Top)$, $f^{-1} (U)$ is open. [it is clear that that
is a topology].


It is clear that under this topology, $f \in M(Y,X)$ is continuous. Does
the converse hold? ie. if $f: Y \rightarrow X$ is continuous, then is
$f \in M(Y,X)$?



Sheaf condition and representability in the category Top
There are lots of counterexamples. Take a property of functions defined between topological spaces, such as bounded, and define the sheaf of functions which have this property locally. This works in most cases, I think.


Define $F(Y) := \{f : Y \to \mathbb{R} : f \text{ is locally bounded}\}$. Clearly this is a subsheaf of $Maps(-,\mathbb{R})$. Assume $F$ is representable, then $X:=F(pt)=\mathbb{R}$ is the representing object and carries the finest topology, making all the $f \in F(Y)$ continuous. Assume $U$ is a nonempty open subset of $X$, and take $Y$ to be an arbitrary bounded interval around which intersects $U$, endowed with the indiscrete topology, and $f : Y \to X$ the inclusion. Then $f \in F(Y)$, thus $f$ is continuous, meaning $Y \subseteq U$. As we can vary $Y$, we get $U = X$. Thus $X$ carries the indiscrete topology. But then $X$ represents the functor $Maps(Y,\mathbb{R})$, which is larger than $F$.


There is another reason that $F$ is not representable, namely $F$ does not preserves colimits. A function $f : Y/\sim \to \mathbb{R}$, whose composition with $Y \to Y/\sim$ is locally bounded, does not have to be locally bounded. Take $Y = \mathbb{R}$ and collapse $\mathbb{Z}$ to one point, and let the supremum of $\mathbb{R} \to \mathbb{R}$ around $z \in \mathbb{Z}$ be $|z|$, but taking the same values at $z$.


A reasonable question is now: Let $F : Top^{op} \to Sets$ a continuous sheaf. Is then $F$ representable? If the canonical map $F \to Maps(-,F(pt))$ is injective, the solution set condition is fulfilled and we may apply Freyds Representability Theorem.
EDIT: Ok it follows from SAFT.


