nLab retract argument

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 $\mathcal{C}^{\Delta[1]}$, hence a morphism $A \stackrel{g}{\longrightarrow} B$ such that in $\mathcal{C}^{\Delta[1]}$ 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 :

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