I would like to consider three sheaves $\mathcal{C}^0$, $\mathcal{H}$ and $\mathcal{S}$ on $\mathbb{C}$ (endowed with the euclidean topology): the first is the sheaf of continuous $\mathbb{C}$-valued functions, the second is the sheaf of holomorphic functions and the third is the sheaf of "overconvergent analytic functions" of radius $1$, by which I mean the subsheaf of $\mathcal{H}^0$ assigning to each edit: connected $U$ the set

$$ \{f\in\mathcal{H}^0(U)\bigm\vert\forall\;x\in U,\text{ the Taylor series of }f\text{ converges in }B(x,1)\}. $$

By analytic continuation, given an open $U\subseteq \mathbb{C}$, if two sections $s_1,s_2\in \mathcal{S}(U)$ coincide everywhere but possibly in a small ball of radius $\rho<1/2$, they must in fact coincide on all $U$ and hence be equal; but this is false if $s_1,s_2\in\mathcal{C}^0(U)$. To me, it looks as if $\mathcal{S}$ is a sheaf for a finer Grothendieck topology (on the category $\operatorname{Top}(\mathbb{C})$), where we declare a collection of opens to cover $U$ even if their union misses a small ball inside $U$. I must say that my motivation is more to see an example of a Grothendieck topology on the category of opens of a "street-man topological space" (forgetting Zariski, étale, fpqc and the like) rather than understanding something new on analytic functions.

I have been trying hard to define a Grothendieck topology different from the "usual" one given by all topological covers on the category $\operatorname{Top}(\mathbb{C})$ and got nowhere: a bestiary of my missed trials is (I refer to this Wikipedia page for axioms T1) and T2))

  1. Covering sieves of an open $U$ are attached to collections of opens which, when adding to the collection a family of disjoint balls $B(x,1)$ (disjointness is crucial to still have a sensible sheaf theory), become a true cover of $U$: it verifies axiom T1) but not T2);
  2. Same as above but with balls $B(x,\delta)$ with $\delta$ smaller than the diameter of $U$: it verifies T2) but not T1);
  3. Covering sieves of $U$ are attached to collections of open balls $B(x,1/2)$ such that the collection $B(x,1)$ covers $U$: it verifies T1) but not T2);
  4. Same as above replacing $1/2$ by $\delta=$diameter of $Y$: verifies T2) but not T1).

I ended up in despair. I'd like either to know if the unique Grothendieck topology on $\operatorname{Top}(\mathbb{C})$ is the "usual one"; or to finally find an example of one a such. Secondly, if possible, I would also like to understand if my idea that analytic continuation can be rephrased in saying that $\mathcal{S}$ is a sheaf for a finer topology than the usual one, makes sense and if it is true or false.

Addendum After a conversation with MO user ACL regarding this question I convinced myself that one can define a topology by declaring that only sieves attached to countable covers (or, I believe, finite covers) are covering sieves. I am unable though to produce an interesting example of a sheaf for either of these topologies which is not a sheaf for the usual one. So I'd be happy with either

  1. new examples of Grothendieck topologies, or
  2. with an example of a "natural" sheaf for one of the two above, or
  3. with something about my original question about $\mathcal{S}$.
  • $\begingroup$ I don't think the edit is enough to solve the problem. What you want is the set of holomorphic functions $f$ such that for every point $x$ the taylor series of $f$ at $x$ at a radius of convergence at least one. What can happen with your current definition is that $U$ is connected but $B(x,1) \cap U$ is not an the taylor serie of $f$ around $x$ converge but differs from the value of $f$ on connected component of $B(x,1) \cap U$ not containing $x$. $\endgroup$ Apr 4, 2016 at 12:07
  • $\begingroup$ @SimonHenry Oh yes, thanks! I'll correct again; any idea about my initial question about topologies? $\endgroup$ Apr 4, 2016 at 12:48

2 Answers 2


There is (as always) a finer topology $T$ making this into a sheaf, the question is whether this topology is different from the usual one or not.

The cover for $T$ of an open subset $V \subset \mathbb{C}$ are the family $V_i \subset V$ such that for any open subset $U \subset \mathbb{C}$, $\mathcal{S}$ satisfies the sheaf conditions with respect to the family $V_i \wedge U \subset V \wedge U$.

Also as it is already a sheaf for the ordinary topology it is enough to look at covering of the form $V' \subset V$ made of a single open subobject: indeed $V_i \subset V$ will be a cover for $T$ if and only if $\bigvee V_i \subset V$ is a cover for $T$.

I claim that for any $z \in \mathbb{C}$, the inclusion $\mathbb{C}-z \subset \mathbb{C}$ is a covering for $T$ (this is basically what Riemann theorem on removable singularity says), so $T$ is indeed strictly finer than the ordinary topology, and in fact the topos of $T$ sheaves is a subtopos of Sh($\mathbb{C}$) with no points.

I initially thought that it would be a sheaf for the double negation topology but it is not the case: $\mathbb{C}-\mathbb{R} \subset \mathbb{C}$ is not a $T$ covering but it is a covering for the double negation topology.

But the converse is true: the double negation topology is finer than $T$, and this follows directly from the fact that for any non zero open subset $U$, $\mathcal{S}(U)$ is not reduced to $0$.

So the additional cover you are going to add when passing from the ordinary topology to $T$ can only be of the form $U \subset U'$ for $U$ a dense open subset of $U'$.

A final note: regarding your question whether the analytic continuation property can be seen as a sheaf condition with respect to a finer topology, I would say: not exactly. The analytic continuation properties corresponds to the concept of a decidable sheaf (see my answer to this question). Decidability is a little related to being separated for the double negation topology but it is a different property.


The subpresheaf $\mathcal S\subseteq\mathcal C^0$ is not a subsheaf.

Indeed, let $U,V\subseteq\mathbb C$ be two disjoint open sets, and consider the function $f:U\cup V\to\mathbb C$ assigning $0$ to points in $U$ and $1$ to points in $V$. Then $f|_U\in\mathcal S(U)$ and $f|_V\in\mathcal S(V)$, but $f\in\mathcal S(U\cup V)$ only if every point in $U$ has distance $\geq 1$ to every point of $V$.

  • 1
    $\begingroup$ Oh yes, sure, thanks. Let me modify my question accordingly— I hope it works. That being said, I would like more to focus on G-tpologies in my question... $\endgroup$ Apr 4, 2016 at 6:01

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