Hello to everyone,

I am studying the properties of combinatorial model categories, following the exposition given by Jacob Lurie in Higher Topos Theory ([HTT] from now on), in section A.2.6.

At some point, he needs to show that in a presentable category $\mathcal C$ and a large enough set $S$ of morphisms in $\mathcal C$, the class generated by $S$ under transfinite pushouts is the same as the class generated by $S$ under retracts and transfinite pushouts; that is: we don't need retracts. This is accomplished in Proposition A.1.5.12.

In the proof of Proposition A.1.5.12, he needs to replace a sequence of morphisms with a tree, satisfying some additional condition; the existence of such a replacement would be Lemma A.1.5.7, but I have problems in understanding why the proof should be correct.

In particular, with the notations used there, I could consider for every $\beta \in A$ the subset $B := \{\alpha \in A \mid \alpha \preceq \beta\}$; this would be $\preceq$-downward closed by construction; since it has a final object, we obtain $$ Y_B^\prime := \varinjlim_{\alpha \in B} Y_\alpha^\prime \simeq Y_\beta^\prime $$ On the other side, condition (1) implies that $B$ has a final object also when thought as subset of $(A,\le)$; it follows that $$ Y_B := \varinjlim_{\alpha \in B} Y_\alpha \simeq Y_\beta $$ i.e. $Y_\beta \simeq Y_\beta^\prime$, so that the diagram shouldn't be changed. But then, I don't see how to prove that $\{Y_\alpha\}_{\alpha \in A'}$ is a $S$-tree (Definition A.1.5.1 in [HTT]).

Therefore, my questions are:

- do you agree with me that the result is seemingly false or can you explain me how the proof is supposed to work?
- do you think that the Proposition A.1.5.12 is correct?
- do you have any other reference for a proposition which is similar to Proposition A.2.6.8 (which is used in the proof of the Smith's characterization of combinatorial model structures)?

**Edit.** I found a related question here. Even though it doesn't answer my question, it fixes the notations I am using, hence I am signaling it for your convenience.