I am reading Lurie's Higher Topos Theory and I need some help to understand a part of the proof of Proposition A.2.6.15 (A.2.6.13 in the published version).

In the proposition, we are working with a locally presentable category $\mathbf{A}$, a class $W$ of morphisms in $\mathcal{C}$, and a set $C_0$ of morphisms in $\mathbf{A}$, subject to the following conditions:

  1. The class $W$ contains all isomorphisms, has the two out of three property, is close under filtered colimits, and has a set $W_0$ such that every morphism in $W$ is a filtered colimit of morphisms in $W_0$.

  2. Given cocartesian squares

$$\require{AMScd} \begin{CD} X @>>> X' @>{g}>>X''\\ @V{f}VV @VVV @VVV\\ Y @>>> Y' @>>{h}> Y'', \end{CD} $$

if $f\in C_0$ and $g\in W,$ then $h\in W$.

  1. Every map in $\mathbf{A}$ having the right lifting property with respect to $C_0$ lies in $W$.

We then want to show that the condition (2) remains valid if we repalce $C_0$ by the weakly saturated class generated by $C_0.$ Lurie proves this by arguing that the class $P$ of morphisms for which (2) remains true when $C_0$ is replaced by $P$ is weakly saturated, i.e., closed under pushouts, transfinite compositions, and retracts.

I understand that $P$ is closed under pushouts and transfinite compositions: Closure under pushout is obvious. Closure under transfinite transfinite composition is a consequence of the fact that $W$ is closed under filtered colimits. But I don't see why $P$ is closed under retracts.

Here's my failed attempt. Let $f\in P$ and suppose $f'$ is a retract of $f$, so that we have a commutative diagram of the form

$$\require{AMScd} \begin{CD} A' @>{i}>> A @>{r}>>A' \\ @V{f'}VV @V{f}VV @V{f'}VV\\ B' @>>{j}> B @>>{s}> B' . \end{CD} $$

We want to show that given cocartesian squares $$\require{AMScd} \begin{CD} A' @>>> X' @>{g}>>X''\\ @V{f'}VV @VVV @VVV\\ B' @>>> Y' @>>{h}> Y'' \end{CD} $$

with $g\in W$, we have $h\in W$. Well, I can only think of pasting the two diagrams to obtain a commutative diagram of the form

$$\require{AMScd} \begin{CD} A @>{r}>> A' @>>> X' @>{g}>>X''\\ @V{f}VV @V{f'}VV @VVV @VVV\\ B @>>{s}> B' @>>> Y' @>>{h}> Y''. \end{CD} $$

If the rectangle comprised of the left two squares is a pushout, then we are done because $f\in P$. But this isn't the case! For instance, the square

$$\require{AMScd} \begin{CD} {\{0,1\}} @>>> \ast \\ @VVV @VVV\\ {\{0,1,2\}} @>>> \ast\\ \end{CD} $$

in $\mathsf{Set}$ is definitely not a pushout even though the right vertical arrow is a retract of the left vertical one. And, given this, I don't know how we can show that $P$ is closed under retracts... Any help is appreciated. Thanks!

  • $\begingroup$ Skimming through the appendix, I see that A.1.5 discusses sufficienct conditions for being closed under retractions to follow from being closed under transfinite composition. I can't tell if the results of that section apply here, however. $\endgroup$ May 13 at 3:44
  • $\begingroup$ @VladimirSotirov I'm afraid that the resultsin A.1.5 does not seem to be applicable to our situation. (In order to apply the results in that section, we will need to have our $C_0$ to be large enough to contain all the maps (up to isomorphism) between $\kappa$-compact objects in the saturation of $C_0$, where $\kappa$ is an uncountable regular cardinal for which $\mathcal{C}$ is locally $\kappa$-presentable. We are not assuming anything like this in the situation at hand.) Thank you for your attention! $\endgroup$
    – Ken
    May 13 at 3:50

1 Answer 1


This question has been answered by Maxime Ramzi in MO. See there for his answer.


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