This is a rather nice question I got from this user 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,

whenever $f: Y \rightarrow X$ in $M(Y,X)$, and $g : Z \rightarrow Y$ continuous, $f \circ g : Z \rightarrow X$ belongs to $M(Z,X)$.

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 $f \in M(Y,X)$?