Contents

# Contents

## Idea

The “retract argument” is a simple but frequently useful standard lemma in discussion of weak factorization systems. It asserts that if a morphism factors as the composition of two factors such that it is has the left or right lifting property against its second or first factor, respectively, then it is a retract (as an object of the arrow category) of the respective other factor.

The retract argument is frequently used in the verification of the axioms of model category structures.

## Statement

###### Lemma

(retract argument)

Consider a composite morphism

$f \;\colon\; X\stackrel{i}{\longrightarrow} A \stackrel{p}{\longrightarrow} Y \,.$

Then:

1. If $f$ has the left lifting property against $p$, then $f$ is a retract of $i$.

2. If $f$ has the right lifting property against $i$, then $f$ is a retract of $p$.

###### Remark

Here by a retract of a morphism $X \stackrel{f}{\longrightarrow} Y$ in some category $\mathcal{C}$ is meant a retract of $f$ as an object in the arrow category $\mathcal{C}^{\Delta}$, hence a morphism $A \stackrel{g}{\longrightarrow} B$ such that in $\mathcal{C}^{\Delta}$ there is a factorization of the identity on $g$ through $f$

$id_g \;\colon\; g \longrightarrow f \longrightarrow g \,.$

This means equivalently that in $\mathcal{C}$ there is a commuting diagram of the form

$\array{ id_A \colon & A &\longrightarrow& X &\longrightarrow& A \\ & \downarrow^{\mathrlap{g}} && \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{g}} \\ id_B \colon & B &\longrightarrow& Y &\longrightarrow& B } \,.$
###### Proof

(of lemma )

We discuss the first statement, the second is formally dual.

Write the factorization of $f$ as a commuting square of the form

$\array{ X &\stackrel{i}{\longrightarrow}& A \\ {}^{\mathllap{f}}\downarrow && \downarrow^{\mathrlap{p}} \\ Y &= & Y } \,.$

By the assumed lifting property of $f$ against $p$ there exists a diagonal filler $g$ making a commuting diagram of the form

$\array{ X &\stackrel{i}{\longrightarrow}& A \\ {}^{\mathllap{f}}\downarrow &{}^{\mathllap{g}}\nearrow& \downarrow^{\mathrlap{p}} \\ Y &= & Y } \,.$

By rearranging this diagram a little, it is equivalent to

$\array{ & X &=& X \\ & {}^{\mathllap{f}}\downarrow && {}^{\mathllap{i}}\downarrow \\ id_Y \colon & Y &\underset{g}{\longrightarrow}& A &\underset{p}{\longrightarrow}& Y } \,.$

Completing this to the right, this yields a diagram exhibiting the required retract according to remark :

$\array{ id_X \colon & X &=& X &=& X \\ & {}^{\mathllap{f}}\downarrow && {}^{\mathllap{i}}\downarrow && {}^{\mathllap{f}}\downarrow \\ id_Y \colon & Y &\underset{g}{\longrightarrow}& A &\underset{p}{\longrightarrow}& Y } \,.$

Last revised on March 22, 2016 at 04:32:59. See the history of this page for a list of all contributions to it.