# Understanding filtration argument

I'm trying to read the proof of Theorem 3.1 in Cellular Categories and Stable Independence. In the proof ii) $$\implies$$ iii), the author uses filtrations to proceed with the argument. I'm having some trouble understanding why the conclusion holds, where the author claims we can express an arrow as a certain transfinite composition. Below I lay some definitions to talk about the proof.

Def: We call an object filtrable if it can be expressed as the colimit of a chain of objects of lower presentability rank (the smallest presentability cardinal), and such chain also has a length below the presentability rank.

Def: A cellular category is a pair $$(\mathcal{K}, \mathcal{M})$$ where $$\mathcal{K}$$ is a cocomplete category and $$\mathcal{M}$$ a set of morphisms, called cellular, containing all isomorphisms and closed under pushouts and transfinite compositions.

We can define a subcategory $$\mathcal{K}_\mathcal{M}$$ of cellular arrows, but this subcategory needn't have pushouts since the induced arrow needn't be cellular. To that end, we define:

Def: A commutative square

is called cellular if the induced arrow from the pushout of the span $$C \leftarrow A \rightarrow B$$, $$P \to D$$, is cellular.

The collection of cellular square is almost a stable independence relation, and the main theorem aims to charaterize when it is in terms of the category being combinatorial.

Def: $$(\mathcal{K}, \mathcal{M})$$ is called combinatorial if $$\mathcal{M} = \operatorname{Cof}(X)$$ (cofibrantly generated), that is, $$\mathcal{M}$$ is the closure under retracts (in the category of arrows), pushouts and transfinite compositions of some set $$X$$.

We have

Theorem 3.1 (part of):
(ii) The cellular squares form an stable independence relation in $$\mathcal{K}_M$$
implies
(iii) $$(\mathcal{K}, \mathcal{M})$$ is combinatorial

From an earlier article it is know that (ii) amounts to the subcategory $$\mathcal{K}_{\mathcal{M}, \downarrow}$$ (which is a subcategory of arrows whose morphisms give cellular squares) being accessible and closed under directed colimits. This allows us to express every object as a retract of an object with a filtration there, and that is how the argument proceeds.

The authors basically take $$\lambda$$ s.t. both $$\mathcal{K}$$, $$\mathcal{K}_{\mathcal{M}, \downarrow}$$ are $$\lambda$$-accessible and almost well-$$\lambda$$-filtrable (every object with rk $$\geq \lambda$$ admits a retract in a filtrable object), then we prove by induction that for every $$\mu$$, $$\mathcal{M}_{\mu^+} \subseteq \operatorname{cof}(\mathcal{M}_\lambda)$$, where $$\mathcal{M}_\mu$$ is the set of arrows in $$\mathcal{M}$$ with $$\mu$$-presentable (co)domain.

The ideia is basically as follows: for any $$h = h_0: K_0 \to L$$ in $$\mathcal{M}_{\mu^+}$$, we WLOG assume $$h_0$$ is filtrable (as $$\operatorname{cof}$$ is retract-closed), then take a filtration

then take a pushout to obtain $$h_1$$:

Noting that by induction hypothesis and pushout-closure, $$h_{0,1} \in \operatorname{cof}(\mathcal{M}_\lambda)$$.
Now, we take a filtration (?) for $$h_1$$ above $$t_{01}$$ (this just seems to be presentability factoring),

then we repeat the same thing we did, but taking a pushout along $$t_{1, 1}$$ instead. This proccess is the successor proccess in transfinite induction, while the limit proccess is just a colimit (on the $$h_{i, i+1}$$). Doing this, we obtain a $$\operatorname{cf}{(\mu)}$$-chain. Then, since $$h = \operatorname{colim}{ h_{i, i+1}}$$, and $$\operatorname{cof}(\mathcal{M}_\lambda)$$ is closed under transfinite inductions, $$h$$ is there too.

First of all,how can $$h_1$$ be filtrable just like $$h_0$$? It seemed that $$h_1$$ was actually supposed to be a retract of such filtrable object, but I don't know if this goes through and why the author would bother with $$h_0$$ being filtrable in first place.

Most importantly, I can't understand how the conclusion is true, and maybe that helps with the retract part. Of course it makes intuitive sense that $$h$$ is the contructed transfinite composition, but either I'm missing some cofinality argument with the $$t_{0i}$$ inclusions and the fact we take pushouts at $$t_{i, i}$$ or something else. I've tried drawing diagrams but really can't get anything substantial to work, the arrows $$t_{0i}$$ are vertically chained while the $$h_{i, i + 1}$$ are horizontally chained, so I can't see how to apply the fact we have $$t_{0, i} \to t_{i, i} \to h_{i, i + 1}$$ in the construction. Does universality of the pushouts induces universality of our cocone $$(h_i: K_i \to L)$$? If so, how?

I'd appreciate if someone could assist me in understanding the proof construction, both the filtration assumption on $$h_1$$ and the overall construction. It needn't be a full blown explanation of the details, just a hint to work out the conclusion, that is, that $$K_{\operatorname{cf}{(\mu)}} = L, h_0 = \operatorname{colim}{ h_{i, i+ 1} }$$.

Any help is appreciated!

EDIT: Transcribed proof above, which anyone can find in pages 9-10 of the linked article.

I guess I understand now how the construction implicitly goes. The ideia is that pushouts induce colimits (hence why we have $$h_1$$ filtrable), and that's how we proceed with the argument. When taking the first pushout, we get induced colimit diagrams
We basically glue these $$t^\prime_{i, j}$$ diagrams together filling the previous spots with identities. That is, at the $$i$$th row we start the arrow $$t^\prime_{i, j}$$ diagram, filling the previous rows with identities. This gives us the following diagram (with compatible cocones which are colimits)
Then $$\operatorname{colim}{D} \implies \operatorname{colim}{c}$$ is precisely