Recall that a presheaf $\mathcal F$ on a topological space $X$ is said to satisfy the "identity theorem" if the following holds:
If $U\subset X$ is a domain and $f,g\in \mathcal F(U)$ are sections which induce the same germ in a single stalk, i.e. if there exists some $x\in U$ such that $\rho_x(f)=\rho_x(g)$, where $\rho_x$ assigns the germ at $x$, then $f=g$.
One can prove that if $X$ is locally Hausdorff and $\mathcal F$ satisfies the identity theorem, then the étalé space $\lvert \cal F \rvert$ is Hausdorff. I'm interested in something like a converse to this result. Therefore, I tried to prove (in the affirmative) the following:
Question: Let $\mathcal F$ be a presheaf on $X$ such that $\lvert \cal F\rvert$ is Hausdorff. Then does $\mathcal F$ satisfy the identity theorem?
The clearest way to approach this is through the use of germs, and indeed one can prove that $S=\{x\in X\mid \rho_x(f)=\rho_x(g)\}$ is clopen (openness is always true, closedness folllows from the Hausdorff assumption on $\lvert \cal F\rvert$), where $f,g$ are sections of $\cal F$ on the domain $U$. Hence $f$ and $g$ induce the same germs in every point. But this is not enough: If the "locality axiom" fails for this presheaf (i.e. the sections are not determined by their restrictions to smaller open subsets that cover $U$) then we can not conclude that $f=g$. Is there a way to work around this, or does one really need to add the locality axiom?
Consider a Riemann surface $X$. The presheaf $\mathcal F$ given by $\mathcal F(U)=\mathcal O^*(U)/\exp(\mathcal O(U))$ on an open subset $U\subset X$ then yields a counterexample.
Indeed, since locally every non-vanishing holomorphic function admits a holomorphic logarithm, the stalks of this presheaf consist of a single element at every point. On the other hand, if one considered an open set biholomorphic to an open, punctured disk $U$ in $\Bbb C$, then the identity map on this punctured disk is not equal to the constant section $1_U\in \mathcal O^*(U)/\exp(\mathcal O(U))$, but they must clearly induce identical elements in each stalk.