It can be shown (see Is every paracompact, Hausdorff, locally contractible space homotopy equivalent to a CW complex?) that if $X$ is a locally contractible paracompact Hausdorff space such that the $\infty$-topos $Sh(X)$ of sheaves on $X$ is hypercomplete then $X$ is homotopy equivalent to a CW complex. On the other hand, if $X$ is actually a CW complex then $X$ is a locally contractible paracompact Hausdorff space. However, it is not clear if in this case $Sh(X)$ is necessarily hypercomplete (it is true, for example, if $X$ is finite dimensional). Note that we are talking about the $\infty$-topos of all sheaves, not just the locally constant ones. The question then naturally arises - is $Sh(X)$ hypercomplete for every CW complex $X$?

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    $\begingroup$ I have heard this asserted. $\endgroup$ Commented Jun 1, 2014 at 21:47
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    $\begingroup$ I removed my answer since I misread something in Higher Algebra. $\endgroup$ Commented Sep 19, 2015 at 5:45

1 Answer 1


ETA The answer is yes in general. Replace 2 below with a reference to HTT, Prop.

Since this has been open for a while, let me give a partial answer which hopefully is already interesting: I believe that the ∞-topos of sheaves on any locally finite (equivalently, locally compact) CW complex $X$ is hypercomplete. Since we know this is true if $X$ is finite-dimensional, the following two observations imply the result:

  1. Any colimit of hypercomplete ∞-topoi is hypercomplete.

  2. If a locally compact space $X$ is a filtered union of closed subspaces $X_i$, then the ∞-topos $\operatorname{Shv}(X)$ is the colimit of the ∞-topoi $\operatorname{Shv}(X_i)$. (Here "locally compact" can be in the sense of Higher Topos Theory, Def.

One might hope to answer the question for arbitrary CW complexes by generalizing 2.

Assertion 1 follows from the fact that hypercomplete ∞-topoi form a coreflective subcategory of ∞-topoi (HTT, Prop.

Assertion 2 can be proved using the description of sheaves on locally compact spaces in terms of K-sheaves (HTT, Def., Cor. I claim that, if $i: Y\hookrightarrow X$ is a closed embedding of locally compact spaces, then precomposition with $i$ preserves K-sheaves.

The only nontrivial point to check is the colimit condition: if $F$ is a sheaf on $X$ and $K\subset Y$ is compact, we must show that $$F(K)=\operatorname{colim}_{K\Subset_Y K'} F(K').$$ The LHS is the colimit of $F(U)$ over the filtered poset of opens $U\subset X$ containing $K$, while the RHS is the colimit of $F(U)$ over the filtered poset of pairs $K'\subset U$ with $K\Subset_Y K'$ and $U$ open in $X$. Since $X$ is locally compact and $Y$ is closed in $X$, it is clear that the "forget $K'$" map between these posets is cofinal.

Thus, the pullback functor $i^*: \operatorname{Shv}(X) \to \operatorname{Shv}(Y)$ is simply given by $(i^*F)(K) = F(i(K))$. Since any compact subset of $X$ belongs to $X_i$ for some $i$ (by a standard argument), we see that a K-sheaf on $X$ is the same thing as a compatible family of K-sheaves on the subspaces $X_i$, which is exactly what Assertion 2 is saying (colimits of ∞-topoi being computed as limits of underlying ∞-categories).

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    $\begingroup$ I just saw the ETA, that's fantastic! too bad I can't vote up twice. $\endgroup$ Commented Mar 27, 2017 at 17:31

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