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,

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

  • $\begingroup$ In the elementary version of your question you assume that $F$ is a subfunctor of a representable functor $Hom(-,X)$. I don't think that this comes for free. $\endgroup$ Jan 30, 2010 at 22:11
  • $\begingroup$ The second version can be explained even to undergraduates. Also even that one is intractable for me. $\endgroup$
    – Anweshi
    Jan 30, 2010 at 23:16
  • $\begingroup$ ok put $X=F(pt)$, then there is a canonical natural map $F(Y) \to Maps(Y,X)$, but I don't see why it is injective. $\endgroup$ Jan 30, 2010 at 23:40

1 Answer 1


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.


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