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Stream: learning: questions

Topic: Pushout complements in Cat

Bryce Clarke (Apr 26 2023 at 15:43):

Suppose I have a fully faithful functor $f \colon A \rightarrow B$. Let $i_{A} \colon A_{0} \rightarrow A$ denote the inclusion of the discrete category of objects. Does this pair of functors always admit a [[pushout complement]]?

I think the pushout complement $P$ should have the same objects as $B$, such that $P(fa, fa') = \emptyset$ if $a \neq a'$, $P(fa, fa) = \{ \ast \}$, and $P(x, y) = B(x, y)$ otherwise. Then the functor $A_{0} \rightarrow P$ should be fully faithful (which is necessary for its pushout along $i_{A}$ to be fully faithful), and the functor $P \rightarrow B$ is identity-on-objects.

Bryce Clarke (Apr 26 2023 at 16:12):

Thinking more about this, I believe we also need to assume that $f$ is injective-on-objects. Otherwise we have a counterexample where $A$ is the [[walking isomorphism]] and $B$ is the terminal category.

Mike Shulman (Apr 26 2023 at 16:41):

I think the pushout of those functors will be the disjoint union of $A$ with $B\setminus A$. You're missing all the morphisms from objects in the image of $A$ to those not in its image, and vice versa.

Bryce Clarke (Apr 26 2023 at 16:50):

Mike Shulman said:

You're missing all the morphisms from objects in the image of $A$ to those not in its image, and vice versa.

Right. When I write $P(x, y) = B(x, y)$, I mean that at most one of x or y is in the image of $A$. In particular, I mean that $P(fa, y) = B(fa, y)$ and $P(x, fa) = B(x, fa)$ for $a \in A$ and $x, y \notin im(f)$.

Bryce Clarke (Apr 26 2023 at 16:51):

The otherwise was doing a lot of work in the first statement and was a bit unclear.

Mike Shulman (Apr 26 2023 at 16:53):

Oh, I see. But then you can't compose a morphism $fa \to y$ with a morphism $y \to fa'$.

Bryce Clarke (Apr 26 2023 at 17:00):

Ah perfect, so that gives a counter-example. Take the functor $\{ 0 < 2 \} \rightarrow \{ 0 < 1 < 2 \}$. Then the ''pushout complement'' that I try to construct above doesn't work.

Bryce Clarke (Apr 26 2023 at 17:31):

I am curious now as to what (neccesary and) sufficient conditions on $f \colon A \rightarrow B$ are such that its pushout complement with $i_{A} \colon A_{0} \rightarrow A$ exists.
It's tempting to add the condition [[Conduche functor]] to the above assumptions, but this feels more like ``plugging the hole'' in the above counter-example rather than actually getting to the bottom of the problem.

Mike Shulman (Apr 26 2023 at 17:35):

Actually that's an interesting question: what does it mean for a full subcategory to be a Conduche functor? A full subcategory is a fibration iff it's a [[sieve]], and an opfibration iff it's a cosieve, but I haven't thought about fully faithful Conduche functors before.

Mike Shulman (Apr 26 2023 at 17:37):

However, I don't think that will work either, because in that case the span $P \leftarrow A_0 \to A$ still doesn't contain any information about the composites of the arrows in $A$ with the arrows not in $A$, so it can't be expected to give back $B$.

Mike Shulman (Apr 26 2023 at 17:39):

Take $\{ 0 < 1 \} \to \{0 < 1 < 2\}$: then $P$ has $0<2$ and $1<2$ but not $0<1$, so the pushout will have two arrows $0\to 2$, one coming from $P$ and one being the freely generated composite of the $0<1$ in $A$ with the $1<2$ in $P$.

Mike Shulman (Apr 26 2023 at 17:40):

Looking at that, I'm tempted to guess the answer is that it only holds if $A$ is complemented, $B = A \sqcup (B\setminus A)$.

Bryce Clarke (Apr 26 2023 at 17:42):

Mike Shulman said:

Take $\{ 0 < 1 \} \to \{0 < 1 < 2\}$: then $P$ has $0<2$ and $1<2$ but not $0<1$, so the pushout will have two arrows $0\to 2$, one coming from $P$ and one being the freely generated composite of the $0<1$ in $A$ with the $1<2$ in $P$.

So for that example, I think that the pushout complement $P$ should be the category $\{0\} \sqcup \{1 < 2 \}$, but that doesn't agree with the construction I attempted to describe above.

Mike Shulman (Apr 26 2023 at 17:43):

Hmm, yes, that does seem to give a pushout complement in that case.

Bryce Clarke (Apr 26 2023 at 17:47):

Mike Shulman said:

Looking at that, I'm tempted to guess the answer is that it only holds if $A$ is complemented, $B = A \sqcup (B\setminus A)$.

Just checking: what is the category $(B \setminus A)$? Is it just the full subcategory of $B$ determined by the objects not in the image of $f \colon A \rightarrow B$?

Mike Shulman (Apr 26 2023 at 17:48):

Yes, that's what I meant.

Mike Shulman (Apr 26 2023 at 17:49):

Let's see, let $f:A\hookrightarrow B$ be a full subcategory inclusion that is a Conduche functor, and $B'$ the full subcategory of $B$ on the complement of $A$. Then Conduche-ness of $f$ means that no object of $B'$ can be both the codomain of an arrow whose domain is in $A$ and the domain of an arrow whose codomain is in $A$. So we can partition the objects of $B'$ into $B_0$, which have an arrow to $A$, $B_1$ which have an arrow from $A$, and $B_2$ which have neither.

Mike Shulman (Apr 26 2023 at 17:51):

There can be no arrows from $B_1$ to $B_0$, from $B_1$ to $B_2$, or from $B_2$ to $B_0$. So $B$ admits a functor to the walking commutative square, with fibers $B_0$, $A$, $B_2$, and $B_1$.

Mike Shulman (Apr 26 2023 at 17:56):

Given the categories $B'$ and $A$, the additional data consists of two profunctors $H_1 : B_1 ⇸ A$, $H_2:A ⇸ B_0$, and a morphism $\alpha : H_2 \circ H_1 \to \hom_B$.

Mike Shulman (Apr 26 2023 at 17:57):

Whereas the data of $P$ consists of, in addition to $B'$, profunctors $K_1 : B_1 ⇸ ob(A)$, $K_2:ob(A) ⇸ B_0$, and a morphism $\beta : K_2 \circ K_1 \to \hom_B$.

Mike Shulman (Apr 26 2023 at 18:00):

There's a forgetful functor $(H_1,H_2,\alpha) \mapsto (K_1,K_2,\beta)$, and taking the pushout should be its left adjoint. So I figure the condition on $f$ -- which is admittedly not a very nice one -- is that $(H_1,H_2,\alpha)$ is in the image of this left adjoint.

Bryce Clarke (Apr 26 2023 at 18:52):

Thanks, I think I am following (although I think that $B_0$ and $B_1$ should be swapped in the profunctors $H_i$ and $K_i$?).
So the forgetful functor is given by some pasting of $\alpha$ with 2-cells in $Prof$ using the representable profunctors of $i_{A} \colon ob(A) \rightarrow A$? (I think this should be straightforward, I just can't "see" it when I write it down).

Bryce Clarke (Apr 26 2023 at 19:20):

I wonder if things simplify somewhat if we assume that $H_1$ or $H_2$ is the initial profunctor (or that $B_0$ or $B_1$ is empty).

Bryce Clarke (Apr 26 2023 at 20:57):

Hmmm, I guess this is obvious but if the pushout complement of $(i_{A} \colon A_{0} \rightarrow A, f \colon A \rightarrow B)$ exists, it must also be a pullback square.

Mike Shulman (Apr 26 2023 at 21:29):

Maybe your conventions about profunctors are the opposite of mine? (-:

Mike Shulman (Apr 26 2023 at 21:29):

For me a profunctor $A⇸B$ is a functor $B^{\rm op}\times A \to \rm Set$.

Mike Shulman (Apr 26 2023 at 21:33):

Bryce Clarke said:

So the forgetful functor is given by some pasting of $\alpha$ with 2-cells in $Prof$ using the representable profunctors of $i_{A} \colon ob(A) \rightarrow A$?

Yes. It may be easier to see in the double category Prof than in the bicategory Prof, where instead of representable profunctors you can use vertical arrows: tikzcd.

Bryce Clarke (Apr 27 2023 at 07:02):

Mike Shulman said:

Whereas the data of $P$ consists of, in addition to $B'$, profunctors $K_1 : B_1 ⇸ ob(A)$, $K_2:ob(A) ⇸ B_0$, and a morphism $\beta : K_2 \circ K_1 \to \hom_B$.

So the idea is that $K_1 : B_1 ⇸ ob(A)$ is the subset of $B(fa, b)$ determined by the morphisms which do not arise from the composition with a non-identity arrow in the image of $A$?

Bryce Clarke (Apr 27 2023 at 07:20):

Mike Shulman said:

There's a forgetful functor $(H_1,H_2,\alpha) \mapsto (K_1,K_2,\beta)$, and taking the pushout should be its left adjoint. So I figure the condition on $f$ -- which is admittedly not a very nice one -- is that $(H_1,H_2,\alpha)$ is in the image of this left adjoint.

I guess I am failing to come up with an example where this condition does not hold; that is, a full subcategory inclusion $f \colon A \rightarrow B$ which is a Conduche functor and does not admit a pushout complement with the inclusion $i_{A} \colon A_{0} \rightarrow A$. I will keep thinking about it.

Bryce Clarke (Apr 27 2023 at 08:00):

Bryce Clarke said:

I am curious now as to what (neccesary and) sufficient conditions on $f \colon A \rightarrow B$ are such that its pushout complement with $i_{A} \colon A_{0} \rightarrow A$ exists.
It's tempting to add the condition [[Conduche functor]] to the above assumptions, but this feels more like ``plugging the hole'' in the above counter-example rather than actually getting to the bottom of the problem.

I should mention that it is not necessary for a full subcategory inclusion $f \colon A \rightarrow B$ to be a Conduche functor for its pushout complement along $i_{A} \colon A_{0} \rightarrow A$ to exist. For example, take $A$ to be the terminal category, $B$ to be the free standing section-retraction pair, and $f$ to pick out the initial object.

So I guess the question is whether it is a sufficient condition.

Mike Shulman (Apr 27 2023 at 15:39):

More generally, whenever $A$ is discrete, $i_A$ is an isomorphism, so the pushout complement always exists.

Mike Shulman (Apr 27 2023 at 15:40):

Bryce Clarke said:

So the idea is that $K_1 : B_1 ⇸ ob(A)$ is the subset of $B(fa, b)$ determined by the morphisms which do not arise from the composition with a non-identity arrow in the image of $A$?

Well, it's more general than that. You would want to take $(K_1,K_2,\beta)$ to be some "freely generating subset" of $(H_1,H_2,\alpha)$.