Categories in which isomorphism of stalks does not imply isomorphism of sheaves

Question

Let $\mathcal{A}$ be a locally small category with colimits of small filtered diagrams. For the purposes of this question, an $\mathcal{A}$-presheaf on a topological space $X$ is a functor $\Omega (X)^\textrm{op} \to \mathcal{A}$, where $\Omega (X)$ is the poset of open subspaces of $X$, and an $\mathcal{A}$-sheaf is an $\mathcal{A}$-presheaf $A$ such that, for all objects $T$ in $\mathcal{A}$, $U \mapsto \mathcal{A} (T, A (U))$ is a sheaf (of sets). The stalk $A_x$ of an $\mathcal{A}$-presheaf $A$ at a point $x$ is the colimit $\varinjlim_U A (U)$ where $U$ varies over the filter of open neighbourhoods of $x$.

Say $\mathcal{A}$ is sheaf-local on a topological space $X$ if, for every morphism $f : A \to B$ of $\mathcal{A}$-sheaves on $X$, if $f_x : A_x \to B_x$ is an isomorphism in $\mathcal{A}$ for all points $x$ in $X$, then $f$ is an isomorphism (of $\mathcal{A}$-sheaves on $X$). It is well known that common categories such as $\textbf{Set}$, $\textbf{Ab}$, $\textbf{CRing}$ and so on are sheaf-local on every topological space. In fact, this is so for any finitely accessible category. It is also clear that sheaf-locality on a discrete space is automatic.

Question. What are some examples of $\mathcal{A}$ and $X$ where $\mathcal{A}$ is not sheaf-local on $X$?

If we generalise to (∞, 1)-categories, then topological spaces that do not have enough points in the ∞-topos sense would provide examples. But I wonder if this phenomenon already happens at the 1-category level.

(This is somewhat related to my old question, is there a good general definition of "sheaves with value in a category"?)

Answer 1

One tends to run into non-conservativity of stalks when working with sheaves of categories with colimits. There are some remarks in the literature about this point, for sheaves valued in the $\infty$-category of presentable stable $\infty$-categories, see for instance https://arxiv.org/pdf/1704.04291 remark 2.8.2.

The $\infty$ is not really essential: we can adapt all that to get an example of a 1-category $\mathcal{A}$ which does the job. Namely, let $\mathcal{A}$ be the category whose objects are small posets with all joins, and whose morphisms are order preserving maps which commute with joins. It then turns out that for sheaves on $\mathbb{R}$ valued in $\mathcal{A}$, taking stalks does not detect isomorphisms.

Let $\mathcal{F}$ be the $\mathcal{A}$-valued sheaf on $\mathbb{R}$ which assigns to each open set $U$ the poset of open subsets of $U$, where for each pair of opens $U \subseteq V$ the restriction $i_{U, V}^*: \mathcal{F}(V) \rightarrow \mathcal{F}(U)$ is given by intersection with $U$. Let $\mathcal{G}$ be the subsheaf of $\mathcal{F}$ which sends each open set $U$ to the poset of open subsets of $U$ which are unions of connected components of $U$. I claim that the inclusion $\mathcal{G} \rightarrow \mathcal{F}$ induces an isomorphism in stalks.

Fix a point $x$ in $\mathbb{R}$. Then the stalk of $\mathcal{G}$ at $x$ can be computed as the colimit of $\mathcal{G}((x-\varepsilon, x+\varepsilon))$ over all $\varepsilon > 0$. This is a constant diagram with value given by the poset $\lbrace 0 < 1 \rbrace$, so the stalk $\mathcal{G}_x$ is given by $\lbrace 0 < 1 \rbrace$.

General results about colimits of locally presentable categories imply that the stalk of $\mathcal{F}$ at $x$ can be computed as the limit $\lim_{U} \mathcal{F}(U)$, where once again we index over all neighborhoods of $x$, but where for each inclusion $U \subseteq V$ the transition in the diagram is given by the morphism of posets $(i_{U, V})_*: \mathcal{F}(U) \rightarrow \mathcal{F}(V)$ which is right adjoint to $i_{U, V}^*$. The maps $(i_{U, V})_*$ are all fully faithful, and hence the stalk of $\mathcal{F}$ at $x$ can be computed as the intersection of the images of $(i_{U, \mathbb{R}})_*$ over all neighborhoods $U$ of $x$. An open subset $W$ of $\mathbb{R}$ in the image of $(i_{U, \mathbb{R}})_*$ necessarily contains the complement of $U$, so we see that the intersection of all these images consists of at most two elements, namely $\mathbb{R}$ and $\mathbb{R}-\lbrace x \rbrace$. Both are realized: for any $U$ we have that $\mathbb{R} = (i_{U, \mathbb{R}})_*(U)$, while $\mathbb{R} - \lbrace x \rbrace = (i_{U, \mathbb{R}})_*(U - \lbrace x \rbrace)$. It follows that the stalk of $\mathcal{F}$ at $x$ is also given by the poset $\lbrace 0 < 1 \rbrace$.

We now have that the stalk of both $\mathcal{G}$ and $\mathcal{F}$ at $x$ is given by $\lbrace 0 < 1 \rbrace$. The morphism $\mathcal{G}_x \rightarrow \mathcal{F}_x$ preserves minimum elements since it is a morphism in $\mathcal{A}$. The only thing left to observe is that the morphism $\mathcal{G}_x \rightarrow \mathcal{F}_x$ is not constant. If it were, one would be able to conclude that the maps $\mathcal{F}(U) \rightarrow \mathcal{F}_x$ are all constant, since they would send both $U$ and the empty set to the same element. But this cannot happen, since the image of the maps $\mathcal{F}(U) \rightarrow \mathcal{F}_x$ should generate the target under joins.

This particular $\mathcal{A}$ is not locally presentable, however one can also find a locally presentable example by working with the category $\mathcal{A}'$ of small posets with countable joins and countable join preserving morphisms. The morphism of $\mathcal{A}$-valued sheaves $\mathcal{G} \rightarrow \mathcal{F}$ from above can also be regarded as a morphism of $\mathcal{A}'$ valued sheaves, and it turns out that the stalks in this setting are still given by $\lbrace 0 , 1 \rbrace$: this can be deduced from the case of $\mathcal{A}$ by using the fact that for every $U$ the posets $\mathcal{F}(U)$ and $\mathcal{G}(U)$ are equal to their own $\operatorname{Ind}_{\omega_1}$-completion.

Answer 2

Here is a reformulation/generalisation of G. Stefanich's counterexample, showing that sheaf-locality can fail very dramatically once we leave the realm of locally finitely presentable categories. More precisely:

Proposition.

1. There is an algebraic category $\mathcal{A}$ that (simultaneously) fails to be sheaf-local on every locally connected Hausdorff space $X$ with at least one open subset that is not closed. ($\mathcal{A}$ does not depend on $X$.)

2. For every uncountable regular cardinal $\kappa$, there is a $\kappa$-ary algebraic category $\mathcal{A}$ that (simultaneously) fails to be sheaf-local on every locally connected Hausdorff space $X$ of cardinality $< \kappa$ with at least one subset that is not closed. ($\mathcal{A}$ depends on $\kappa$ but not $X$.)

Proof. Recall that the Sierpiński space $S$ is the topological space whose set of points is $\{ 0, 1 \}$ and whose open subsets are $\emptyset, \{ 1 \}, \{ 0, 1 \}$. Almost by construction, there is a natural bijection between continuous maps $X \to S$ and open subsets of $X$, so we have a sheaf $\Omega$ on $X$ whose sections over an open $U \subseteq X$ are open subsets of $U$.

Similarly, there is a natural bijection between continuous maps $X \to 2$ and clopen subsets of $X$, so we have a sheaf $Q$ on $X$ whose sections over an open $U \subseteq X$ are the clopen subsets of $U$. Furthermore, the continuous injective map $2 \to S$ gives rise to a subsheaf inclusion $Q \to \Omega$.

Now let $\mathcal{A}$ be the category of $\kappa$-ary join semilattices, i.e. posets in which every subset of cardinality $< \kappa$ has a least upper bound, and assume $\kappa$ is greater than the cardinality of $X$. This is a $\kappa$-ary algebraic category (hence locally $\kappa$-presentable a fortiori). Although $\mathcal{A}$ has filtered colimits, they are not preserved by the forgetful functor $\mathcal{A} \to \textbf{Set}$ if $\kappa > \aleph_0$. This is the root cause of the failure of sheaf-locality for $\mathcal{A}$.

Let $N (x)$ be the filter of open neighbourhoods of $x$ in $X$. There is a natural morphism $\Omega (U) \to \Omega (1)$ in $\mathcal{A}$ sending $V$ to $1$ if and only if $x \in V$, so we get a morphism $\varinjlim_{U : N (x)^\textrm{op}} \Omega (U) \to \Omega (1)$. If $X$ is Hausdorff, this is an isomorphism: indeed, if $V \in \Omega (X)$ and $x \notin V$, then $V$ is the union of $< \kappa$ open $V_\alpha \subseteq V$ such that $x \notin \overline{V_\alpha} \subseteq X$; but if $V' \in \Omega (X)$ and $x \notin \overline{V'}$, then we can find $U \in N (x)$ with $U \cap V' = \emptyset$, so $V'$ must be identified with $\emptyset$ in the colimit, and hence $V$ is also identified with $\emptyset$ in the colimit. (This argument would also work when we consider $\Omega$ as a sheaf of sets if $N (x)^\textrm{op}$ is $\kappa$-filtered.)

On the other hand, if $X$ is locally connected, then it is easy to see that $\varinjlim_{U : N (x)^\textrm{op}} Q (U) \cong Q (1)$, being a colimit of a cofinally constant diagram. Thus, if $X$ is a locally connected Hausdorff space with at least one open subset that is not closed, then $Q_x \to \Omega_x$ is an isomorphism for every $x \in X$, but $Q \to \Omega$ is not an isomorphism.

The above proves the second claim of the proposition. For the first claim, remove the cardinality restrictions and replace $\mathcal{A}$ with the category of complete join semilattices.　◼