TITLE: Associativity of pullbacks
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: 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 $R \to A\times_D(B\times_E C)$.