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Stream: community: general

Topic: Characterizing the poset reflection of presheaves

Andrej Bauer (Dec 09 2023 at 18:06):

Let [C,D] be the functor category from C to D. Let 2 be the poset 0 → 1.

Is there a known characterization of the poset reflection of [C, Set] in terms of C? Specifically, is it [r(C), 2] for some poset r(C) derived from C?

Examples (these were given by Amir Akbar Tabatabai from Utrecht in personal communication):

1. The poset reflection of Set is 2, so r(1) = 1.
2. The poset reflection of [T, Set] where T is a tree (qua poset) is [T, 2], so for a tree-like-poset r(T) = T.
3. Given a group G qua category, the poset reflection of [G, Set] is [Sub(G), 2] where Sub(G) is the post of subgroups of G, ordered by H ≤ K ⇔ ∃ g ∈ G . g⁻¹ H g ⊆ K.

Morgan Rogers (he/him) (Dec 09 2023 at 19:22):

That's a fun puzzle, let's figure it out. I'll fix a category $C$ and refer to the poset reflection of $[C,\mathrm{Set}]$ as $P$.

My first observation is that the poset of subterminal objects of $[C,\mathrm{Set}]$ is a subposet of $P$. That's isomorphic to the poset of upward-closed subsets of the poset reflection of $C$ (in other words, subsets of the set of objects of $C$ such that if $a$ is in the subset and there exists a morphism $a \to b$ then $b$ is in the subset too). These include the minimal and maximal element of $P$.

Morgan Rogers (he/him) (Dec 09 2023 at 19:22):

Any object of $[C,\mathrm{Set}]$ has a support, which is subterminal object obtained from the epi-mono factorization of the unique map to the terminal object. If the map from any object to its support splits, then $P$ is isomorphic to the poset of subterminals. This is the case in your example for trees, and it also holds whenever $C$ is a poset in which each upward-closed subset is a disjoint union of upward closed sets generated by a single element (like... a disjoint union of trees?)

Morgan Rogers (he/him) (Dec 09 2023 at 19:24):

How can that support map fail to split? You've already seen that it doesn't split for $C$ a group, or indeed a monoid, but let's stick with $C$ a poset for now, and more specifically the cospan $a \rightarrow c \leftarrow b$. The diagrams $1 \rightarrow 1 \leftarrow 0$ and $0 \rightarrow 1 \leftarrow 1$ are subterminal objects in $[C,\mathrm{Set}]$ represented by $a$ and $b$, respectively. Their coproduct, which is a diagram of coproduct inclusions $1 \rightarrow 2 \leftarrow 1$ has the terminal object $1 \rightarrow 1 \leftarrow 1$ as its support but there is no section to that map.

Morgan Rogers (he/him) (Dec 09 2023 at 19:33):

If I take any cospan in $F:C \to \mathrm{Set}$, then either:

• $F(a)$ is empty, so the support is (a subobject of) the diagram represented by $b$ and the map splits
• $F(b)$ is empty, so the reverse happens, or
• Both $F(a)$ and $F(b)$ are inhabited, in which case the support is the terminal object and splits unless all of the elements in $F(a)$ and $F(b)$ are mapped to disjoint elements in $F(c)$, in which case the map to the support factors through the coproduct inclusion diagram described above and the factoring map splits.

Morgan Rogers (he/him) (Dec 09 2023 at 19:36):

In this case $P$ can be identified with the collection of upward closed sets of the 4-element 'diamond' poset... which looks rather like a product or finite limit completion of $C$. Let's see if that insight generalizes.

Morgan Rogers (he/him) (Dec 09 2023 at 19:43):

I'll switch back to generic $C$.
A further observation is that $C^{\mathrm{op}}$ embeds into $[C,\mathrm{Set}]$ (for this point it's a shame that you chose covariant functors rather than presheaves, but let's continue!)
Thus $P$ admits a functor from the poset reflection of $C^{\mathrm{op}}$. More strongly, the support of a diagram $F$ is $\{c \in C \mid F(c) \neq \emptyset\}$, which is precisely the set of $c$ such that $F$ admits a morphism from the diagram represented by $c$. So in $P$, (the image of) $F$ is bounded below by the (image of the) representables and bounded above by the union of their supports.

Morgan Rogers (he/him) (Dec 09 2023 at 20:23):

Now let's go back to the group case, so that there is exactly one representable to think about. Actually, I prefer monoids, since the fact that it's a group isn't obviously important (yet). I'll call the representable diagram $M$ (that's the monoid acting on its own set of elements).
Consider the poset of quotients of $M$, which we can identify with the set of left congruences; for groups, the equivalence classes split $M$ into cosets of the class containing $1$, which is a subgroup. As you established, for a group, we get maps in both directions between quotients if and only if those subgroups are conjugate. More generally, for a congruence $R$ on $M$ and element $m$, we can construct $(R)^*m = \{(x,y) \mid (xm,ym) \in R\}$, and maps from $M/R$ to $M/R'$ correspond to elements $m$ such that $R \subseteq (R')^*m$. Note that this poset can be degenerate: any monoid with an absorbing element will have a global section, for example.

So we extract the poset reflection of this category of congruences (pedantic note that you described a preorder of subgroups). From here, we can glue together these principal actions to get new elements of $P$. For a group, all the quotients will appear as independent orbits in each action, and we can always identify identical orbits, so we do indeed get the upward-closed subsets.
For a general monoid, it seems like things should be mildly more interesting. Consider $\N$. This acts on $\Z$ but there is no transformation $\Z \to \N$ which respects the action.

Morgan Rogers (he/him) (Dec 09 2023 at 21:51):

(I may continue tomorrow, but in case anyone wants to join in and figure out what the poset reflection is for $[\N,\mathrm{Set}]$, do go ahead!)

Todd Trimble (Dec 10 2023 at 04:45):

This topic is connected with the theory of exact completions, and the question of when the exact completion $E_{ex}$ of a category $E$ is a topos. For the connection, one should consult the work of Menni, for example here just to give one point of entry, which gives a lot of relevant references.

Todd Trimble (Dec 10 2023 at 04:45):

If $y: E \to E_{ex}$ denotes the embedding of a topos into its exact completion, then it's been mentioned by Carboni that the lattice of subobjects of $y(X)$ is isomorphic to the poset reflection of $E/X$, and even for some simple presheaf toposes $E$, this need not be small. He mentions that $E =$ the topos of quivers $\mathsf{Set}^{\rightrightarrows}$ is an example where this happens: its posetal reflection is not small.

Todd Trimble (Dec 10 2023 at 04:45):

Deep, deep within the bowels of the nLab, there is a footnote in the article on regular and exact completions, which explains further. It reads:

Todd Trimble (Dec 10 2023 at 04:45):

"For example, let $C = Set^{\bullet \rightrightarrows \bullet}$ be the topos of directed graphs. For each ordinal $\alpha$, let $G_\alpha$ be the directed graph whose nodes are elements of $\alpha$ and with a directed edge from $\beta$ to $\gamma$ if $\beta \lt \gamma$ in $\alpha$. Then in the poset reflection $Pos(C)$, we have a class of proper monomorphisms, e.g., $[G_\alpha] < [G_{\alpha'}]$ whenever $\alpha < \alpha'$. Thus $Pos(C)$ is a large poset. This example also shows that $Pos(C)$ need not be a total category even if $C$ is."

Andrej Bauer (Dec 10 2023 at 10:38):

Thanks, the connection with exact completions is very useful to know. I suppose the observation about quivers tells us not to expect a very nice solution?

Morgan Rogers (he/him) (Dec 10 2023 at 11:50):

If a category as simple as the parallel pair gives a proper class (note that the same construction of an irreflexive quiver from a poset means we get at least the poset reflection of the category of posets and injective monotone functions) then I suspect that there is little chance of realising your original conjecture @Andrej Bauer . That said, you may at least be able to prove that this reflection has the properties you would expect, like small joins.