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Stream: learning: questions

Topic: smallest image subcategory

David Egolf (Jan 10 2022 at 00:59):

I have started reading "Generalized Congruences - Epimorphisms in $\mathsf{Cat}$" as part of an effort to understand questions related to those raised in in #learning: questions > equivalence relationships preserved by endofunctors.

The authors note that, given a functor $F: A \to B$, the image of $F$ is not generally a category. However, they note it is possible to create a "smallest" subcategory of $B$ containing the objects and morphisms of $F(A)$.

The "smallest" of anything usually seems to be a limit or colimit of some kind. To get practice with this kind of thinking, I was trying to express "smallest subcategory of $B$ containing the image $F(A)$" as a limit or colimit. However, it feels like I am running into a problem based on the fact that $F(A)$ isn't necessarily a category. In particular, I'm not seeing how to diagrammatically describe a category $C$ that contains the objects and morphisms of $F(A)$.

Any hints are appreciated.

Zhen Lin Low (Jan 10 2022 at 01:01):

"Just" take the class of all subcategories containing the image and take their intersection (= greatest lower bound).

David Egolf (Jan 10 2022 at 01:04):

That would work, thanks.
However, I guess I was hoping to describe the property "is a subcategory containing the image $F(A)$" using a diagram. My reasoning is that things described as diagrams are often easier to generalize or dualize.

Zhen Lin Low (Jan 10 2022 at 01:05):

That would be a subcategory through which F factors. You could express it as a commutative triangle.

David Egolf (Jan 10 2022 at 01:09):

Ah, I think that's it!
If I understand you correctly, category $C$ would be such a category:
subcategory containing image

where $i$ is a monomorphism and the diagram commutes.

Zhen Lin Low (Jan 10 2022 at 01:28):

Yes. The image (for want of a better name) would be the smallest (= initial) such category.

John Baez (Jan 10 2022 at 01:46):

David wrote:

The authors note that, given a functor F:A→B, the image of F is not generally a category.

Why not?

(Please: nobody else answer this.)

David Egolf (Jan 10 2022 at 02:28):

Intuitively, one reason this happens is because pairs of morphisms that weren't compatible to compose in $A$ can become compatible to compose in $B$. By a pair of morphisms that are "compatible to compose", I mean two morphisms where the source of one is the target of the other. This can result in a situation where we can't compose some compatible morphisms in $F(A)$. If this happens, then $F(A)$ fails to be a category.

In more detail, say $m_1$ and $m_2$ are two non-compatible morphisms in $A$. Let $F: A \to B$ act so that $F(m_1)$ and $F(m_2)$ become compatible to compose. For $F(A)$ to be a category, we need to be able to compose these to form $F(m_1) \circ F(m_2)$. If $m_1$ and $m_2$ were compatible in $A$, then we would have $F(m_1 \circ m_2) = F(m_1) \circ F(m_2)$. That, is we would be able to form this composite in the image of $F$. However, since $m_1$ and $m_2$ aren't compatible, there is no guarantee we can form their composite in the image of $F$.

The authors of that paper have a nice figure illustrating this:
image of G is not a category

In the category on the right, the composite of the two morphisms in the middle is the morphism on the right. However, $G$ sends $h$ to a different morphism. $G(f)$ and $G(g)$ are compatible to compose, but are not actually possible to compose while staying within the image of $G$. So, the image of $G$ is not a category here.

David Egolf (Jan 10 2022 at 02:31):

Incidentally, I suppose this particular problem only occurs when $A$ has more than one object. So, unless there are other problems I'm not thinking of, we can expect the image of a functor from a category with one object to be a category. I suppose this is one way to see that the image of a group homomorphism has at least a chance to be a subgroup (should be closed under composition).

John Baez (Jan 10 2022 at 03:41):

Nice! I don't get why the picture involves two functors, $F$ and $G$. We were talking about a single functor $F$. So that picture is confusing to me. But it doesn't matter much, since your verbal explanation before the picture is fine.

Saying it in my own way: to check that the image of $F$ is a subcategory we need to check that the composite of two composable morphisms in the image, say $F(f): a \to b$ and $F(g) : b \to c$, is again in the image. That is, we need to check that $F(g)F(f)$ is $F$ of some morphism. But what could it be? The only thing to guess here is $F(gf)$, but that may not make sense, since the target of $f$ may not be the source of $g$. Once you notice this you can easily cook up a counterexample.

David Egolf (Jan 10 2022 at 03:45):

(The reason there are two functors in the picture is that the picture is a screenshot from that paper, and they're illustrating something else with it).

John Baez (Jan 10 2022 at 04:18):

Okay. So I like this stuff a lot, and it makes want to give you a little puzzle.

What's the simplest condition you can put on the functor $F$ to avoid the problem we just noticed?

John Baez (Jan 10 2022 at 04:18):

For a functor with this property, its image will be a subcategory.

David Egolf (Jan 10 2022 at 04:44):

Hmmm, let's see.
We want any two morphisms in the image of $F: A \to B$ to be composable into a morphism that is still in the image of $F$. This will let the image of $F$ be a category.
So, for any composable morphisms $F(u): S \to T$ and $F(v): T \to V$ in the image of $F$ (where $u,v$ are morphisms in $A$) we need $F(v) \circ F(u)$ to also be in the image of $F$. That is, we need $F(v) \circ F(u) = F(w)$ for some morphism $w$ in $A$. I think this is all we need.

John Baez (Jan 10 2022 at 05:17):

That's true, but that's just basically saying that the image of $F$ is a subcategory.

John Baez (Jan 10 2022 at 05:18):

I was looking for a simple interesting condition on $F$ that implies the image of $F$ is a subcategory. Maybe it takes some mathematical expertise to see this condition, so I'll tell you what I'm talking about: "$F$ is one-to-one on objects".

David Egolf (Jan 10 2022 at 05:24):

Oh, interesting! I guess the idea is that morphisms that aren't composible stay non-composable in the image in this case. That's because the only way to make morphisms composable when they weren't composable to begin with is to send objects that were different to objects that are the same: so that sources and targets that were distinct become the same.

John Baez (Jan 10 2022 at 05:25):

Right.

John Baez (Jan 10 2022 at 05:27):

There are functors that aren't one-to-one on objects whose image is still a subcategory, but the "obvious" way to show a composite $F(g)F(f)$ is in the image of $F$ is to have it equal $F(gf)$, and if $F$ is one-to-one on objects that will always be true.

Oscar Cunningham (Jan 10 2022 at 13:48):

Demanding that the functor was full would also do it.

Zhen Lin Low (Jan 10 2022 at 14:16):

It would suffice for the functor to "reflect isomorphy" in the sense that objects that become isomorphic in the codomain must already be isomorphic. This is not a very natural condition though. Full-on-isomorphisms would be a better categorification of "injective map".