representable functors - 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.