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?

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.

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    $\begingroup$ Consider the constant presheaf $P(U) = 2$, where $2$ is a set containing 2 elements, on the Cantor space. The sheafification will be $P(U) = \{f : U \to 2 \mid f$ continuous$\}$. Since the Cantor space is totally disconnected, $P(U)$ will always be infinite. $\endgroup$ 2 days ago
  • $\begingroup$ @MarkSaving I see, thank you for the explanation! However, I had thought the sheafification of the constant presheaf is the "locally constant" sheaf? For the case of the Cantor set, does the locally constant sheaf agree with the "continuous function" sheaf you described? $\endgroup$ 2 days ago
  • $\begingroup$ Edit: I've changed the question to focus on describing the conditions on $X$ for the above property to hold. $\endgroup$ 2 days ago
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    $\begingroup$ The locally constant sheaf is exactly the sheaf I described. A function being locally constant is the same as it being continuous (if $2$ has the discrete topology - I guess I should have specified this). $\endgroup$ 2 days ago
  • $\begingroup$ @MarkSaving I see, thanks! $\endgroup$ 2 days ago

2 Answers 2


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 $n\in\mathbb{N}$.)

  • $\begingroup$ I would add the property can hold for a non-Alexandrov space. Note that this property depends only on the locale. Consider the space $\mathbb{Z} \cup \{-\infty\}$ with basic opens $[-\infty, n]$ for $n \in \mathbb{Z}$. This space is not Alexandrov, but its locale of opens is the same as that of its subspace $\mathbb{Z}$, which is Alexandrov. $\endgroup$ 2 days ago
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    $\begingroup$ I mentioned almost that exact counterexample in my latest edit :) $\endgroup$ 2 days ago
  • $\begingroup$ It's actually not obvious that this property depends only on the locale, though. The open sets that satisfy (1) depend only on the locale, but whether for every point all sufficiently small neighborhoods satisfy (1) would seem to depend on what points you have. $\endgroup$ 2 days ago
  • $\begingroup$ Your new answer seems a bit mixed up. You are discussing the claim for all $x$, there exist arbitrary small open sets $x \in U$ such that for all $F$, $(\phi_F)_U$ is an isomorphism. But the question discusses the claim that for all $x$, for all $F$, there exist small open $x \in U$ such that $(\phi_F)_U$ is an isomorphism. $\endgroup$ 2 days ago
  • $\begingroup$ As for why the property depends only on the locale: consider some presheaf $F$. The claim in the language of spaces is that for all $x$, for all neighbourhoods $U$ of $x$, there exists some open neighbourhood $V \subseteq U$ of $x$ such that $(\phi_F)_V$ is an isomorphism. An easy rephrasing of this claim in a point-free manner is that for all open $U$, the set $\{V \subseteq U \mid V$ open, $(\phi_F)_V$ is an isomorphism$\}$ is an open cover of $U$. $\endgroup$ 2 days ago

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.


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