Sheaf with values on $\mathcal C$ not the same as a presheaf with values in $\mathcal C$ which is a sheaf of sets

Question

Let $\mathcal C$ be a category where all products exist. Let $X$ be a topological space, and let $\textrm{Ouv}_X$ be the category of open sets of $X$. A presheaf with values in $\mathcal C$ is a contravariant functor $F: \textrm{Ouv}_X \rightarrow \mathcal C$. Let $U$ be an open set in $X$, with open cover $U_{\alpha}$. We have an obvious morphism $r: F(U) \rightarrow \prod\limits_{\alpha} F(U_{\alpha})$.

We also have two morphisms $r', r'': \prod\limits_{\alpha} F(U_{\alpha}) \rightarrow \prod\limits_{\alpha, \beta} F(U_{\alpha} \cap U_{\beta})$ defined as follows: for $r'$, we have for each $\alpha$ a morphism $\prod\limits_{\alpha} F(U_{\alpha}) \rightarrow F(U_{\alpha})$, and having fixed $\alpha$, a further collection of morphisms $F(U_{\alpha}) \rightarrow F(U_{\alpha} \cap U_{\beta})$ for each $\beta$. This defines a collection of morphisms $\prod\limits_{\alpha} F(U_{\alpha}) \rightarrow F(U_{\alpha} \cap U_{\beta})$ over all $(\alpha, \beta)$, whence the morphism $r'$. We define $r''$ in the same way, but take $F(U_{\beta} \cap U_{\alpha})$ instead of $F(U_{\alpha} \cap U_{\beta})$ (thus indexing over all $(\beta,\alpha)$).

If $r$ is always an equalizer of the morphisms $r', r''$, then $F$ is called a sheaf.

On the other hand, if $\mathcal C$ is a concrete category, then we can compose $F$ with the forgetful functor $| \cdot |$ to obtain a sheaf of sets.

Is the converse true? If $F$ is a presheaf which is a sheaf when regarded as being in the category of sets, is $F$ a sheaf?

Answer 1

If the forgetful functor $\mathcal{C}\to Set$ reflects limits, then $F$ will always be a sheaf if $|F|$ is a sheaf. So to find a counterexample, you will need to work in a concrete category for which the forgetful functor fails to reflect limits, which rules out most of the familiar algebraic categories.

As you suggest, then, we can use $\mathcal{C}=Top$. Let $A$ and $B$ be your favorite non-discrete spaces, and let $C$ be $A\times B$ with the discrete topology. We can then define a $\mathcal{C}$-values presheaf on $\{0,1\}$ by $F(\{0\})=A$, $F(\{1\})=B$, and $F(\{0,1\})=C$ (and $F(\emptyset)=\{*\}$) with the restriction maps being the projections $C\to A$ and $C\to B$. This gives a sheaf of sets, but not a sheaf of topological spaces, since the map $C\to A\times B$ is not a homeomorphism.

Answer 2

I have a very stupid example which I think works. Let $X$ be a topological space, and for each open set $U$, let $F(U)$ be the set of functions from $U$ into $\mathbb{C}$. We can identify $F(U)$ with $\prod\limits_{x \in U} \mathbb{C}$ and give it the product topology. Except I will arbitrarily declare $F(X)$ to be given the discrete topology. Then for a typical open cover $U_{\alpha}$ of $X$,

$$F(X) \rightarrow \prod\limits_{\alpha} F(U_{\alpha})$$

is not going to be a homeomorphism onto its image.