# How to verify sheaf condition in this example?

I am learning about sheaves of sets on a site with a subcanonical topology and have a question.

$f:A\rightarrow Hom(-,X)$ is a map from a pre-sheaf $A$ (for which I want to verify sheaf condition) into a representable sheaf. There is a covering (from a covering family in the site) of $Hom(-,X)$ by $U_j\rightarrow Hom(-,X)$, $j\in J$ with $U_j\times_{Hom(-,X)}A$ is a sheaf for every $j$. Can I conclude that $A$ is a sheaf?

(Actually I have $U_j\times_{Hom(-,X)}A\cong U_i\times Hom(-,Y)$ with one $Y$ for all $j$ but it perhaps suffices to know that it is a sheaf)

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Are we talking about sheaves of sets on a site here? If so, you may want to add the topos-theory tag and clarify your question. – Zhen Lin May 30 '12 at 9:57
Hi Zhen Lin! What need clarification? – sheafstudent May 30 '12 at 10:07
I still don't really understand the setting you're working in. Is $U_j$ itself a representable sheaf, and by cover, do you mean a covering family in the site? You seem to be implicitly assuming a subcanonical topology. Or is $U_j$ an arbitrary sheaf, and by cover do you mean a family which is jointly epimorphic in the sheaf topos? (Also, is $Y$ fixed in advance, or does it vary depending on $U_j$?) – Zhen Lin May 30 '12 at 10:13
Hi Zhen Lin! Is the question ok now? – sheafstudent May 30 '12 at 10:19
OK, I think I'm beginning to understand what you're really asking about. I'm afraid the answer to both your questions is no – I'll post a counterexample if you want. – Zhen Lin May 30 '12 at 10:33

Let $\{ f_\alpha : U_\alpha \to X \}$ be a non-trivial covering family in a subcanonical site $(\mathbb{C}, J)$. By non-trivial, I mean one which does not contain a split epimorphism. We may assume without loss of generality that $\mathbb{C}$ contains a terminal object, since we may always freely add such an object.
Consider the corresponding family $\{ H_{f_\alpha} : H_{U_\alpha} \to H_X \}$ in the presheaf topos $\hat{\mathbb{C}} = [ \mathbb{C}^\textrm{op}, \textbf{Set} ]$. Let $\mathfrak{U}$ be the joint (presheaf!) image of this family; explicitly, $$\mathfrak{U}(C) = \{ f \in \mathbb{C}(C, X) : f \text{ factors through some } f_\alpha \}$$ i.e. $\mathfrak{U}$ is the sieve generated by $\{ f_\alpha : U_\alpha \to X \}$. Since we assumed $\{ f_\alpha : U_\alpha \to X \}$ is not a trivial cover, $\mathfrak{U}$ is a strict subobject of $H_X$. It is clear by construction that $H_{U_\alpha} \cong H_{U_\alpha} \times_{H_X} \mathfrak{U}$, and each of these is a sheaf because $J$ is a subcanonical topology. Moreover, if $Y$ is the terminal object of $\mathbb{C}$, then $H_{U_\alpha} \cong H_{U_\alpha} \times H_Y$, since $H_Y$ is terminal in $\hat{\mathbb{C}}$.
I claim that $\mathfrak{U}$ is not a $J$-sheaf. Indeed, there is a canonical matching family for $\{ f_\alpha : U_\alpha \to X \}$ in $\mathfrak{U}$, namely $\{ f_\alpha : U_\alpha \to X \}$. Since $\mathfrak{U}$ is a subpresheaf of the $J$-sheaf $H_X$, any amalgamation of this matching family would have to be $\textrm{id}_X : X \to X$, yet $\textrm{id}_X \notin \mathfrak{U} (X)$.
That's a very nice example. I am grateful to you. Is it correct that your $\mathfrak{U}$ is a non-separated presheaf? Perhaps for a separated presheaf, being a sheaf is a local condition? – sheafstudent May 30 '12 at 11:30
Any subpresheaf of a sheaf is automatically separated, so in particular $\mathfrak{U}$ is separated. – Zhen Lin May 30 '12 at 11:34