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:
-
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 $W_0$.
-
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 $h\in W$.
- 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!
Best Answer
This question has been answered by Maxime Ramzi in MO. See there for his answer.