Consider two arrows in a category: $k:A\longrightarrow D$ and $h:B\longrightarrow D$ and take their pullback $$\begin{array} AA\times_DB & \stackrel{k'}{\longrightarrow} & B \\ \downarrow{h'} & & \downarrow{h} \\ A & \stackrel{k}{\longrightarrow} & D \end{array} $$ Then consider two other arrows: $g:B\longrightarrow E$ and $f:C\longrightarrow E$ and take the following pullback $$\begin{array} ((A\times_DB)\times_EC & \stackrel{(gk')'}{\longrightarrow} & C \\ \downarrow{f'} & & \downarrow{f} \\ A\times_DB & \stackrel{gk'}{\longrightarrow} & E \end{array} $$ Now, using the same arrows, consider first the following pullback $$\begin{array} BB\times_EC & \stackrel{g''}{\longrightarrow} & C \\ \downarrow{f''} & & \downarrow{f} \\ B & \stackrel{g}{\longrightarrow} & E \end{array} $$ and then the following $$\begin{array} AA\times_D(B\times_EC) & \stackrel{k''}{\longrightarrow} & B\times_EC \\ \downarrow{(hf'')''} & & \downarrow{hf''} \\ A & \stackrel{k}{\longrightarrow} & D \end{array} $$ What I want is an isomorphism $$(A\times_DB)\times_EC\simeq A\times_D(B\times_EC)$$ My guess is that the two sides of that relation are two pullbacks of the same object, hence isomorphic, but I can't prove it. More precisely, I think we can define a pullback of three (or more) objects, then show that the two sides of that relation are both "realizations" of the ternary pullback of $A,B$ and $C$, hence isomorphic.

This is what I call the associativity property of pullbacks and it is my first question. Looking for references, I found on Borceux categorical algebra handbook vol 1, the following: the associativity of pullbacks is defined as the following property (which I can understand and prove): consider a commutative diagram $$\begin{array} AA & \stackrel{f}{\longrightarrow} & B & \stackrel{g}{\longrightarrow} & C\\ \downarrow{h} & & \downarrow{k}& & \downarrow{l} \\ D & \stackrel{p}{\longrightarrow} & E & \stackrel{q}{\longrightarrow} & F \end{array} $$ Then the following facts hold true:

1) if both squares are pullbacks then the outer rectangle is a pullback

2) if the right square and the outer rectangle are pullbacks, then the left square is a pullback

So my second question is: why this property is called associativity of pullbacks? Is there a relation between this notion and the one I described above?

  • 2
    $\begingroup$ Regarding your first question: (a) just show that in either case you have constructed the limit of the diagram $A\to D\leftarrow B \to E \leftarrow C$ (in fact you only need to show this once, since the situation is symmetric). (b) using the Yoneda embedding it is sufficient to prove this in the category of sets. $\endgroup$
    – Nex
    Dec 7, 2016 at 5:16

1 Answer 1


What you are calling associativity and what they are calling are associativity are related as follows. Consider the diagram $$\require{AMScd}\begin{CD} R @>{r_2}>> B\times_E C @>{q_2}>> C\\ @V{r_1}VV @VV{q_1}V @VV{f}V\\ A\times_D B @>{p_2}>> B @>{g}>> E\\ @V{p_1}VV @VV{h}V \\ A @>>{k}> D \end{CD}$$ where each of the squares are pullbacks. Using 1 of from Borceux's categorical algebra horizontally and vertically we see that $$\require{AMScd}\begin{CD} R @>{q_2r_2}>> C\\ @V{r_1}VV @VV{f}V\\ A\times_D B @>>{gp_2}> E \end{CD}$$ and $$\require{AMScd}\begin{CD} R @>{r_2}>> B\times_E C \\ @V{p_1r_1}VV @VV{hq_1}V\\ A @>>{k}> D \end{CD}$$ are pullbacks. The desired isomorphism is then obtained as the composite of the (obvious) isomorphisms $(A\times_D B)\times_E C \to R$ and $R \to A\times_D(B\times_E C)$.


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