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:

  1. do you agree with me that the result is seemingly false or can you explain me how the proof is supposed to work?
  2. do you think that the Proposition A.1.5.12 is correct?
  3. 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.


1 Answer 1


Looks like a typo. Condition $(4)$ should say that $B$ is downward closed under $\leq$, not under $\preceq$ (otherwise, $Y_B$ is not defined).

  • 1
    $\begingroup$ Now I see. Thank you very much for your answer! $\endgroup$ Jun 14, 2013 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.