For what base spaces is the sheafification adjunction unit always an isomorphism on sufficiently small opens?

Question

Let $X$ be a topological space. We use this as the base space for all presheaves below.

Since sheafification $F \mapsto \tilde{F}$ is left adjoint to the inclusion of sheaves into presheaves, this gives a natural presheaf morphism $\phi_F : F \rightarrow \tilde{F}$, which is the unit of the aforementioned adjunction.

Let us say $F$ has property $\mathscr{P}$ if for every $x \in X$ the map $(\phi_F)_U : F(U) \rightarrow \tilde{F}(U)$ is an isomorphism for sufficiently small open nbhds $U$ of $x$.

I was wondering, if $X$ is e.g. a topological manifold then does every presheaf on $X$ have property $\mathscr{P}$? What other kinds of conditions on $X$ are necessary/sufficient?

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Edit: changed the focus of the question to describing the case when $\phi_F$ is an isomorphism, instead of a counter-example. Thank you to the comment below giving a counter-example with $X$ the Cantor set.

Answer 1

A $T_1$ space $X$ has your property iff it is discrete. Discrete spaces trivially have your property (just take $U$ to be $\{x\}$); conversely, suppose $X$ is not discrete. Let $x\in X$ be such that $\{x\}$ is not open. Define a presheaf $F$ on $X$ as follows: $F(U)$ is the power set of $U$ if $x\in U$ and $F(U)$ is a singleton otherwise. If $x\in U\subseteq V$, the restriction map $F(V)\to F(U)$ is $W\mapsto W\cap U$. Let $U$ be any open neighborhood of $x$. Since $\{x\}$ is not open, there is a point $y\neq x$ in $U$. Now I claim that $U$ and $U\setminus\{y\}$ are two elements of $F(U)$ that map to the same section of the sheafification. Indeed, their restrictions to $U\setminus\{y\}$ are equal, as are their restrictions to $U\setminus\{x\}$, and these two sets form an open cover of $U$. Thus $F(U)\to\tilde{F}(U)$ is not injective.

(I'm not sure what can be said in general for non-$T_1$ spaces. Your property holds more generally for Alexandrov spaces, since if $U$ is the smallest open neighborhood of a point $x\in X$ then it is not hard to see that $F(U)\to\tilde{F}(U)$ is always an isomorphism. There are also non-Alexandrov spaces with your property, for instance $\mathbb{N}\cup\{\infty\}$ with the topology consisting of the empty set and sets of the form $[n,\infty]$ for $n\in\mathbb{N}$.)

Answer 2

The simplest example of a manifold is $\mathbb{R}$ itself. This also provides us with a simple counterexample.

Consider the sheaf $F(U) = \{f : U \to \mathbb{R} \mid f$ continuous and bounded$\}$. Note that $F$ is a dense sub-presheaf of $G(U) = \{f : U \to \mathbb{R} \mid f$ continuous$\}$; therefore, $G$ is the sheafification of $F$, and the inclusion map is $\phi_F$ (up to unique isomorphism, of course).

But for no $U$ save the empty set is $(\phi_F)_U$ a surjection. In other words, for $U$ non-empty, there is a continuous $f : U \to \mathbb{R}$ which is not bounded. To see this, suppose $U$ is non-empty. If $U$ is bounded above, consider $f(x) = \frac{1}{(\sup U) - x}$, which is an unbounded continuous function. And if $U$ is not bounded above, $f(x) = x$ is a continuous unbounded function.