Let $C$ be a category endowed with a Grothendieck topology $J$ of covering sieves, giving us a site $(C,J)$. We say $(C,J)$ is locally connected if for any object $c\in C$ all covering sieves $j\in J(c)$ of $c$ are connected as full subcategories of the slice category $C/c$. [1]

Recall that a category is connected when its groupoid completion is connected. Note also that they seem to require that the covering sieves are nonempty.

It seems that this condition is always satisfied when the site comes from a topological space. Given any sieve $j\in J(c)$ let $a \to c, b\to c$ be elements from $j\in J(c)$ we may simply take the fiber product (=intersection) $a\times_c b$ which clearly maps to $a,b$ in $C/c$. Hence it seems all sieves are always connected. It is, of course, possible that $a\times_c b$ is the empty set, but this is allowed.

Concretely, the Cantor space $2^{\mathbb{N}}$, considered as a site in the obvious way, should not be locally connected. I was not able to show this using the above definition. How to show this?



In the site corresponding to a topological space, most objects have the property that all covering sieves are connected, for the reason that you explained, but there's an exception: The empty set is an object that is covered by the empty sieve. (I don't mean the sieve $\{\varnothing\}$, which also covers $\varnothing$; I mean the sieve $\varnothing$.) And this sieve is not connected.

Why not? I ordinarily define a category $C$ to be connected if, whenever it is represented as the disjoint union of a family of categories, the family contains $C$ itself. But the empty category is the disjoint union of the empty family of categories, which doesn't contain the empty category.

The nLab defines connectedness as "$\text{Hom}(C,-)$ preserves finite coproducts." Again, that makes the empty category not connected.

  • $\begingroup$ A hint that $\varnothing$ would be a problem is the nLab page's Remark 3.1, that on a locally connected site, constant presheaves are sheaves. But every sheaf $X$ on a topological space has $X(\varnothing)=1$, whereas most constant presheaves don't have that property. $\endgroup$ – Andreas Blass Apr 26 at 17:39
  • $\begingroup$ Do I understand correctly that this means the site corresponding to a topological space is never locally connected? And that therefore the notions of locally connectedness of a topological space and locally connectedness of the corresponding site are... totally disconnected? $\endgroup$ – merle Apr 26 at 19:32
  • $\begingroup$ @merle As far as I can see, the site of all open subsets of a topological space (with the obvious notion of cover) is never locally connected. But the site of nonempty open sets produces the same (more precisely an equivalent) topos of sheaves, and it can be locally connected. It looks as if that slightly smaller site will be locally connected iff the space is. (Note that nLab only says that local connectedness of a site is sufficient for local connectedness of the topos of sheaves; there's no claim that it's necessary.) $\endgroup$ – Andreas Blass Apr 26 at 19:46
  • $\begingroup$ It seems to me that the site of nonempty open sets corresponding to a disconnected topological space also becomes locally disconnected. Is there a way to fix this? $\endgroup$ – merle Apr 27 at 8:51
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    $\begingroup$ @merle I think you're probably right and I oversimplified things in my previous comment. I don't have the time to check carefully right now, but here's something I think will work: Consider the site whose objects are only the connected open sets (and still using the usual notion of cover). If the space is locally connected, then every open set is covered by connected open subsets, so this site should still give the right topos. And it seems to avoid the problem that you found. $\endgroup$ – Andreas Blass Apr 27 at 14:00

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