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Stream: theory: category theory

Topic: How to define algebraic weak factorization systems

James Deikun (Jun 06 2026 at 04:51):

The normal definition of algebraic weak factorization systems is complex and full of redundant data. It seems like it would be simpler to define an AWFS as a functorial factorization (section of composition) that is skew-coassociative -- that is:

• each (rooted) binary tree induces a higher-arity factorization operation
• binary trees are partially ordered by reachability via left rotations
• the function from binary trees to higher-arity factorization operations extends to a functor.

This seems to lead to the same definition but i'm having a little trouble proving it because of the complexity of the original definition -- making sure I've verified all the laws and there are no extras. Maybe there is a more conceptual approach via the "Beck theorem"?

Nathanael Arkor (Jun 06 2026 at 18:44):

I don't have any insights to lend, but just to mention that Bourke shared an intuition that AWFSs were conceptually similar to skew structures in this talk.

Patrick Nicodemus (Jun 07 2026 at 13:39):

In "Cofibrantly Generated Natural Weak factorization Systems" Garner gives a more conceptual/abstract definition of AWFS as bialgebras in a certain monoidal category, this is described in section 3

Tom Hirschowitz (Jun 08 2026 at 07:27):

I'd be interested in a more hands-on, equivalent definition, but can't make any sense of your candidate. Could you please give a bit more detail?

James Deikun (Jun 09 2026 at 09:41):

Thanks @Patrick Nicodemus -- I haven't done it yet for time reasons, but this definition seems like it would be a lot easier to prove (in)equivalent to mine.

James Deikun (Jun 09 2026 at 09:50):

@Tom Hirschowitz, let's call $[n]$ the category that's just an $n$-arrow path. Then $\mathcal C^{[1]}$ is the arrow category of $\mathcal C$, and the inclusion of $[1]$ in $[n]$ as the longest arrow induces by restriction a standard composition functor $\circ_n : \mathcal{C}^{[n]} \to \mathcal{C}^{[1]}$. Call a section of $\circ_n$ an $n$-ary functorial factorization. The full subcategory of $\mathbf{Cat}(\mathcal{C}^{[1]}, \mathcal{C}^{[n]})$ on functorial factorizations will be known as $\mathbf{Ffac}_n(\mathcal C)$.

James Deikun (Jun 09 2026 at 09:58):

Now, the identity functor is the unique unary functorial factorization, and because of associativity of composition in categories, and the fact that $[n]$ is an $n$-ary pushout of copies of $[1]$, there's an operation where you can "plug" $n$ functorial factorizations $F_i$ into an $n$-ary functorial factorization $G$ to get a functorial factorization $G(F_1,...,F_n)$. This has the effect of factoring an arrow with $G$ and then factoring arrow number $i$ in the path further with $F_i$, then stringing all the resulting paths together. This operation is functorial in $G$ and all the $F_i$.

James Deikun (Jun 09 2026 at 10:13):

So we start with a generating binary factorization $F$ and then we use this operation and the identity along with $F$ to generate all the functorial factorizations we can. Then assume we happen to have a morphism (natural transformation) $\alpha : F(\mathrm{Id},F) \to F(F,\mathrm{Id})$. If $Ff$ as a diagram in $\mathcal C$ looks like $X \xrightarrow{Lf} Ef \xrightarrow{Rf} Y$, then $\alpha_f$ looks like:

$$\begin{CD} X @>Lf>> Ef @>LRf>> ERF @>RRf>> Y \\ @| @V\Delta_fVV @VV\mu_fV @| \\ X @>>LLf> ELf @>>RLf> Ef @>>Rf> Y \\ \end{CD}$$

where the left square will become, in the conventional definition of AWFS, the comultiplication of the comonad $L$ and the right square will become the multiplication of the monad $R$, and the middle square the distributive law between them.

James Deikun (Jun 09 2026 at 11:10):

The next step is to generate all the natural transformations we can by plugging $\alpha$ in various places using the functorial action of ${-}(...)$, and then we want any two parallel arrows we generate freely this way to be equal. It turns out due to various previously-known facts about coherences that this happens exactly when the "pentagon law" holds: $\alpha(F,\mathrm{Id},\mathrm{Id}) \circ \alpha(\mathrm{Id},\mathrm{Id},F) = F(\mathrm{Id},\alpha) \circ \alpha \circ F(\alpha, \mathrm{Id})$. This can be reexpressed pointwise as this hard-to-read diagram in $\mathcal C$.

James Deikun (Jun 09 2026 at 11:16):

The pentagon law is equivalent to three laws, if you can manage to follow the diagram: the comonad comultiplication law for $L$, the monad multiplication law for $R$, and the "only nontrivial law for the distributive law". I think the comonad counit laws for $L$ and the monad unit laws for $R$ are true basically by definition. Is that all the actual laws for an AWFS? It's hard to tell, for me.

Tom Hirschowitz (Jun 10 2026 at 13:15):

This looks cute, thanks for taking the time to write it! I'm not sure I entirely understand the "hard-to-read" diagram though... I'd be interested if you write this up at some point.

James Deikun (Jun 10 2026 at 13:28):

The counit/unit laws turn out to be a little more interesting than I imagined and I'm trying to figure out how best to formulate them.

James Deikun (Jun 10 2026 at 13:31):

(One counit/unit pair states that the diagonal of the middle square of $\alpha_f$ is an identity, which seems like a "nice" law but I'm not sure quite what the significance is; I can't yet interpret the other pair usefully.)