Let $X$ be a topological space, and let $A$ be a presheaf on $X$.

Let $\mathscr{A}$ be the sheaf of germs on $X$. We define a a topology on $\mathscr{A}$ as follows:

Given an open set $U \subset X$, fix a section $s \in A(U)$ and consider the germ $s_x$, of $s$, at $x \in U$. The set of all germs $s_x$ for all $x \in U$ is defined to be open in this topology on $\mathscr{A}$.

In general, the sheaf $\mathscr{A}$ is not Hausdorff. My question is:

Can anyone give an explicit, non-trivial example of a Hausdorff sheaf?


1 Answer 1


The sheaf of germs of holomorphic functions on a complex manifold is Hausdorff. In the same vein, the sheaf of germs of real-analytic functions on a real-analytic manifold. When you have an identity theorem that two functions are globally identical (on a connected open set) if they coincide on a nonempty open set, the sheaf of germs of such functions is Hausdorff.

If the base space is Hausdorff, germs in different stalks can always be separated by disjoint neighbourhoods, so to see the above are Hausdorff, we need only concern ourselves with germs in the same stalk, say above $p$. Let $g_1\neq g_2$ be two germs above $p$ and $s_1, s_2$ two sections on $U_1$ resp $U_2$ representing the germs. Let $U \subset U_1 \cap U_2$ be a connected open neighbourhood of $p$, and $t_1,t_2$ the restrictions of $s_i$ to $U$. Then $[U,t_1] = \{ t_{1x} : x \in U\}$ and $[U,t_2]$ are disjoint neighbourhoods of $g_1$ and $g_2$. For, if there were a germ $c_y \in [U,t_1]\cap [U,t_2]$, that would mean there is a neighbourhood $W \subset U$ of $y$ such that $t_1\lvert_W = t_2\lvert_W$, and by the identity theorem, then $t_1 \equiv t_2$, and thus $g_1 = t_{1p} = t_{2p} = g_2$.

  • $\begingroup$ And can you give an explicit example of a non-Hausdorff sheaf? $\endgroup$
    – Neuromath
    Jul 14, 2019 at 10:12
  • 2
    $\begingroup$ @Neuromath The sheaf of germs of continuous functions on $\mathbb{R}$. The germs of $f \colon x \mapsto x$ and $g \colon x \mapsto \lvert x\rvert$ at $0$ have no disjoint neighbourhoods. For every $y > 0$ we have $f_y = g_y$, and every neighbourhood of either $f_0$ or $g_0$ contains all germs $f_y$ (${} = g_y$) for $0 < y < \varepsilon$ with a suitable $\varepsilon > 0$ (which depends on the neighbourhood, of course). $\endgroup$ Oct 2, 2019 at 16:00

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