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Stream: theory: category theory

Topic: factorisation systems

Morgan Rogers (he/him) (Nov 17 2020 at 09:37):

I've encountered plenty of orthogonal factorisation systems in my time, but I recently came across a result of the form "every morphism in $\mathcal{C}$ factors in a canonical way as an $x$ followed by a $y$", but such that applying this property to the $y$ part could also give a non-trivial factorisation.

Moreover, this particular factorisation construction does extend to an orthogonal factorisation system, but doesn't quite do so in a natural way, because while we can iterate the procedure a countable number of times to reach a fixed point (in the sense that applying the factorisation to the limit will not give a new factorisation), the intermediate morphism when we factor the $y$ part again is non-trivial. Diagrammatically, I end up with:
$A \xrightarrow{p} B \xrightarrow{q} C,$
where $p$ is of type $x$ and $q$ is of type $y$, and such that when I factorize $q$ I get:
$A \xrightarrow{p} B \xrightarrow{p'} B' \xrightarrow{q'} C,$
with $p'$ not an isomorphism, but such that there does exist an isomorphism/equivalence $B \cong B'$ identifying the two factorisations.

My question is: does this situation sound familiar to anyone? Are there any results out there that might be usefully applicable in a situation like this?

Chad Nester (Nov 17 2020 at 09:56):

This feels a lot like computing the normal form of a term in a terminating rewriting system. Like the arrow $t$ that we start with is the term we want to normalise, and then factoring $t$ as $p ; q$ is like saying that via rewrite $p$ we obtained $q$ from $t$, like $t \stackrel{p}{\to} q$. Then if $q$ is in normal form it does not factor further, but if not then we can do another step, and so on.

Morgan Rogers (he/him) (Nov 17 2020 at 10:32):

What's uncomfortable about my situation in that analogy is that you can keep rewriting forever; instead of reaching a normal form, you reach a limit where the rewriting is doing something like adding a term to the end of a countably long expression, so that the resulting expression is equivalent to its predecessor.

Reid Barton (Nov 17 2020 at 11:30):

It sounds a bit like the small object argument. I'm not sure what exactly the orthogonal factorization system is, though. $(x, y)$ can't be an orthogonal factorization system if a map of type $y$ can be factored as a non-isomorphism of type $x$ followed by another map of type $y$.

Morgan Rogers (he/him) (Nov 17 2020 at 12:02):

Hmm so maybe the result isn't orthogonal after all, which wouldn't be that surprising considering the following example.
Consider the following mono-epi (rather than epi-mono) factorisation construction in $\mathbf{Set}$. If I have a morphism $X \to Y$, I can canonically factor it as $X \hookrightarrow X+Y \twoheadrightarrow Y$. Iterating this process up to the countable limit, I get a factorisation $X \hookrightarrow (\omega \times X) + Y \twoheadrightarrow Y$. Doing it one more time, we still get a non-trivial factorisation, but the isomorphism $1+ \omega \cong \omega$ gives a lifting in the resulting square.

Reid Barton (Nov 17 2020 at 12:05):

Yes, this is the factorization you get by running the small object argument on the single map $\varnothing \to *$ for $\omega$ steps.

Morgan Rogers (he/him) (Nov 17 2020 at 12:07):

So your comment was that $x,y$ can't be the actual classes involved in the factorisation system, I see.

Morgan Rogers (he/him) (Nov 17 2020 at 12:08):

And the small object argument only gives a weak factorisation system, so one needs to do some extra work to demonstrate whether uniqueness of lifts holds?

Reid Barton (Nov 17 2020 at 12:28):

Right, or if you want to guarantee up front that you will get an orthogonal factorization system, then for each generating map $A \to B$ you can also add the map $B \amalg_A B \to B$ (one step is enough, because repeating this yields an isomorphism $B \to B$).

Reid Barton (Nov 17 2020 at 12:29):

Or you can check that the maps $B \amalg_A B \to B$ belong to the left class of the wfs that comes out (most likely by building them as cell complexes).

Reid Barton (Nov 17 2020 at 12:31):

I wrote something about this in section 4.6 of https://arxiv.org/abs/2004.12937 but I think it's all very standard.

Morgan Rogers (he/him) (Nov 17 2020 at 13:05):

It may be standard, but it's not an area of the literature I'm so familiar with, so that's very helpful. Thanks Reid!