nLab
retract argument

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

Then:

  1. If ff has the left lifting property against pp, then ff is a retract of ii.

  2. If ff has the right lifting property against ii, then ff is a retract of pp.

Remark

Here by a retract of a morphism XfYX \stackrel{f}{\longrightarrow} Y in some category 𝒞\mathcal{C} is meant a retract of ff as an object in the arrow category 𝒞 Δ[1]\mathcal{C}^{\Delta[1]}, hence a morphism AgBA \stackrel{g}{\longrightarrow} B such that in 𝒞 Δ[1]\mathcal{C}^{\Delta[1]} there is a factorization of the identity on gg through ff

id g:gfg. id_g \;\colon\; g \longrightarrow f \longrightarrow g \,.

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

id A: A X A g f g id B: B Y B. \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 ff as a commuting square of the form

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

By the assumed lifting property of ff against pp there exists a diagonal filler gg making a commuting diagram of the form

X i A f g p Y = Y. \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

X = X f i id Y: Y g A p Y. \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 :

id X: X = X = X f i f id Y: Y g A p Y. \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.