The small object argument

We will talk about the small object argument in this post, that is due to Quillen, and is well suited with respect to the lifting properties. See Section 7.12 in [1].

Let \mathcal{C} be a category admitting small colimits. Suppose B:\mathbb{Z}^+\rightarrow \mathcal{C} is a functor from a small category \mathbb{Z}^+ to \mathcal{C}. Then there is a natural map B(n)\rightarrow \mathrm{colim}~B which induces a map \mathrm{Hom}(A,B(n))\rightarrow\mathrm{Hom}(A,\mathrm{colim}~B) for each object A\in\mathcal{C}. Thus there is a natural map

\underrightarrow{\mathrm{colim}}~\mathrm{Hom}(A,B(n))\rightarrow \mathrm{Hom}(A,\mathrm{colim}~B)        (\ast).

We call A is sequencially small if the natural map (\ast) is an isomorphism.

Recall that in a category \mathcal{C}, a map p:X\rightarrow Y is said to have right lifting property (RLP) with respect to a map i:A\rightarrow B if for every commutative diagram

\displaystyle \begin{matrix}A&\rightarrow&X\\\downarrow^i&&\downarrow^p\\B&\rightarrow&Y\end{matrix}

there is a lifting map B\rightarrow X. Similarly, i is said to have left lifting property (LLP) with respect to the map p.

In the subsequent, given a family of maps \mathcal{F}=\{f_i:A_i\rightarrow B_i~|~i\in I\} in a category \mathcal{C}, we are going to show that there is a way of factorizing the map p:X\rightarrow Y in \mathcal{C} as X\rightarrow X'\rightarrow Y such that X\rightarrow X' has the RLP with respect to every map in \mathcal{F}, provided that each A_i is sequencially small.

Standard construction.

Denote by S^0_i=\{(g,h)~|~pg=hf_i\}.

\displaystyle \begin{matrix}A_i&\stackrel{g}{\rightarrow}&X\\\downarrow^{f_i}&&\downarrow^p\\B_i&\stackrel{h}{\rightarrow}&Y\end{matrix}

Then we define G^1(\mathcal{F},p) as the pushout of the following diagram

\displaystyle \begin{matrix}\coprod\limits_{i,(g,h)\in S^0_i}A_i&\rightarrow&X\\\downarrow^{\coprod f_i}&&\downarrow^{\mu_1}\\\coprod\limits_{i,(g,h)\in S^0_i}B_i&\rightarrow&G^1(\mathcal{F},p)\end{matrix}

thus there is a map p_1:G^1(\mathcal{F},p)\rightarrow Y such that p=p_1\mu_1.

Inductively, suppose G^k(\mathcal{F},p) has constructed with p_k:G^k(\mathcal{F},p) such that p=p_k\mu_k. Denote by S^k_i=\{(g,h)~|~p_kg=hf_i\}. Then we construct G^{k+1}(\mathcal{F},p)=G^1(\mathcal{F},p_k) as the pushout of the following diagram

\displaystyle \begin{matrix}\coprod\limits_{i,(g,h)\in S^k_i}A_i&\rightarrow&X\\\downarrow^{\coprod f_i}&&\downarrow^{\mu_{k+1}}\\\coprod\limits_{i,(g,h)\in S^k_i}B_i&\rightarrow&G^1(\mathcal{F},p_k)\end{matrix}

thus there is a map p_{k+1}:G^{k+1}(\mathcal{F},p)\rightarrow Y such that p=p_{k+1}\mu_{k+1}.

We denote by G^\infty(\mathcal{F},p)=\underrightarrow{\mathrm{colim}}G^n(\mathcal{F},p), p_\infty:G^\infty(\mathcal{F},p)\rightarrow Y and \mu_\infty:X\rightarrow G^\infty(\mathcal{F},p).

Proposition. Suppose each A_i is sequencially small. Then p:X\rightarrow Y factorizes as p=p_\infty\mu_\infty such that p_\infty has the RLP with respect to every map in \mathcal{F}.

Proof. Since each A_i is sequencially small, there is some k such that the map A_i\rightarrow G^\infty(\mathcal{F},p) factors through G^k(\mathcal{F},p). Consider the following diagram

\displaystyle\begin{matrix}A_i&\rightarrow &G^k(\mathcal{F},p)&\rightarrow &G^{k+1}(\mathcal{F},p)&\rightarrow &G^\infty(\mathcal{F},p)\\ \downarrow^{f_i}&&\downarrow^{p_k}&&\downarrow^{p_{k+1}}&&\downarrow^{p_\infty}\\ B_i&\rightarrow&Y&=&Y&=&Y\end{matrix}

Then the pushout construction from G^k(\mathcal{F},p) to G^{k+1}(\mathcal{F},p) implies that there is a map B_i\rightarrow G^{k+1}(\mathcal{F},p) making the diagram commute. In particular, composed with the structure map G^{k+1}(\mathcal{F},p)\rightarrow G^\infty(\mathcal{F},p), we deduce that p_\infty has the desired RLP with respect to the family \mathcal{F} of maps.\square

References

[1] W. G. Dwyer and J. Spalinski, Homotopy theories and model categories, 1995

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