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Disclaimer: This is a cross-listing of a math.stackexchange post. While not research level, after a week of no response, I figured I would ask it here.

For a topological group $G$ and a topological space $X$, denote by $\underline{G^X}$ the sheaf of continuous functions from $X$ into $G$.

Suppose we have an exact sequence of groups $$ 1\rightarrow F\rightarrow G\rightarrow H\rightarrow 1. $$ What are sufficient conditions for the corresponding sequence of sheaves $$ 1\rightarrow \underline{F^X}\rightarrow \underline{G^X}\rightarrow \underline{H^X}\rightarrow 1 $$ to be exact?

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  • $\begingroup$ @Will That sheafification is exact is proved here in the Stacks Project. (Since links in comments are hard to read - it's here: stacks.math.columbia.edu/tag/00WJ) $\endgroup$
    – user62675
    Feb 19, 2015 at 1:40
  • $\begingroup$ @Will Yeah, that's true. $\endgroup$
    – user62675
    Feb 19, 2015 at 1:45
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    $\begingroup$ @Will: I don't understand how you intend to apply that statement to this case. What presheaves are you sheafifying? Note that you are not sheafifying the constant presheaves. $\endgroup$ Feb 19, 2015 at 3:49
  • $\begingroup$ @QiaochuYuan Isn't the case with discrete $G$ solved by sheafifying the constant presheaves? Anyhow, you're definitely right it doesn't give an answer in general. My mistake. I've deleted my comments. $\endgroup$
    – Will Chen
    Feb 19, 2015 at 7:30

2 Answers 2

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The following is perhaps more of an extended comment than an answer. The sequence of sheaves is exact iff the quotient map $G\to H$ has a section over a neighborhood of every point (in fact, because of the group structure, it suffices to have a section over any single nonempty open set). In particular, for instance, this means the sequence of sheaves is always exact if $G$ is a Lie group. In general, however, the sequence need not be exact, as the following example shows.

Let $G$ be the free topological group on the Cantor set $K$; this is just the free group on the underlying set equipped with the obvious colimit topology (in general that topology might fail to be a group topology, but everything works fine because $K$ is compact). Similarly, let $H$ be the free topological group on $[0,1]$. Let $p:K\to[0,1]$ be any continuous surjection, and consider the induced map $q:G\to H$. It is not hard to see this is a quotient map of topological groups. But $q$ cannot have a section over any open set, since $H$ is locally connected and $G$ is totally disconnected.

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Here is a positive answer when $G$ is the universal covering group of $H$ (so then $F=\pi_1H$ which is discrete, and $\pi_1G=0$), such as $\mathbb{Z}\to\mathbb{R}\to S^1$.

Denote by $M(U,G)$ the $G$-valued continuous functions on open $U\subset X$. Then covering space theory says the following induced sequence is exact: $$0\to M(U,F)\to M(U,G)\to M(U,H)\to[U,H]\to 0$$ where $[\cdot,\cdot]$ is homotopy classes of maps. But $Sheaf([\cdot,H])=0$, because given any $f:U\to H$ and $x\in U$ there is a neighborhood $x\in V\subset U$ with $f|_V$ nullhomotopic. Thus your desired sheafification is exact.

In general, for my argument I need a lifting criterion, which means I need the surjection $G\to H$ to be a fibration, moreover with the fiber $F$ discrete (so that I can talk about unique path lifting). Thus this also works for the case of discrete groups.

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  • $\begingroup$ I also expect that in general $G \to H$ being a fibration is necessary. $\endgroup$ Feb 19, 2015 at 3:50
  • $\begingroup$ @QiaochuYuan Indeed, from pg. 161 of Brylinski's Loop spaces, Characteristic Classes and Geometric Quantization: "Since the mapping [$G\rightarrow H$] is a principal $\mathbb{C}^*$-fibration, one has an exact sequence of sheaves of continuous functions . . .". (In this case, $G$ is the group of bounded invertible operators on a separable infinite-dimensional complex Hilbert space and $H$ is the quotient of $G$ by non-zero scalars.) In particular, he seems to suggest that being a fibration is also sufficient, though I known not the relevance of $\mathbb{C}^*$ in particular. $\endgroup$ Feb 19, 2015 at 4:16

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