Proposition 1   If $P$ is a pullback of $f:A\to C$ and $g:B\to C$ , then $P$ is unique (up to isomorphism). This is the justification that $P$ may be written as $A\times_C B$ .

Proof. Let

be the corresponding pullback diagram. First note that if $h:P\to P$ is a morphism such that $p_A=p_A\circ h$ and $p_B=p_B\circ h$ , then $h=1_P$ . This follows from the universal property of pullbacks (for more detail, see the proof of the uniqueness of product here).

Now if $Q$ is another pullback of $f:A\to C$ and $g:B\to C$ with pullback diagram

then there are unique morphisms $x:P\to Q$ and $y:Q\to P$ such that $p_A=q_A\circ x$ , $p_B=q_B\circ x$ and $q_A=p_A\circ y$ , $q_B=p_B\circ y$ . So $p_A=p_A\circ (y\circ x)$ and $p_B = p_B\circ (y\circ x)$ . Therefore, $y\circ x=1_P$ . Similarly $x\circ y=1_Q$ . As a result $P$ is isomorphic to $Q$ .

Proposition 2   Let $I$ be the disjoint union of non-empty sets $J,K$ . Let $I' = \lbrace x_i:C_i\to C \mid i\in I\rbrace$ be a set of morphisms in $\mathcal{C}$ . Let $X,Y$ , and $Z$ be the generalized pullbacks of $I'$ , $J' = \lbrace x_i \mid i\in J\rbrace$ , and $K' = \lbrace x_i \mid i\in K\rbrace$ respectively. Then $X\cong Y\times_C Z$ .

Proof. [Sketch of Proof.] This proof is analogous to the proof of a similar property regarding arbitrary products (see here), so we will skip the diagrams and be brief here. First, note that there are unique morphisms $y:X\to Y$ and $z:X\to Z$ , which results in a unique morphism $f:X\to Y\times_C Z$ . On the other hand, the morphisms $Y\to C_j$ and $Z\to C_k$ give us a well-defined collection of morphisms $Y\times_C Z \to C_i$ for all $i\in I$ , which results in a unique morphism $g: Y\times_C Z \to X$ . There are a number of commutative diagrams relating $f$ and $g$ . In the end, one proves that $g\circ f = 1_C$ and $f\circ g=1_{Y\times_C Z}$ .

Corollary 1 (commutativity of pullbacks)   $X\times_C Y\cong Y\times_C X$ , provided that one of them (and hence the other) exists.

Corollary 2 (associativity of pullbacks)   $X\times_C (Y\times_C Z) \cong X\times_C Y \times_C Z \cong (X\times_C Y) \times_C Z$ , provided that they exist.

Corollary 3   If $\mathcal{C}$ has pullbacks, then it has finite generalized pullbacks.

Proposition 3   Let $\lbrace x_i: C_i\to C\mid i\in I\rbrace$ be a collection of morphisms indexed by a set $I$ . Let $\alpha,\beta:J\to I$ be two surjections. Suppose $D,E$ are the generalized pullbacks of $\lbrace x_{\alpha(j)} \mid j\in J \rbrace$ and $\lbrace x_{\beta(k)}\mid k\in J \rbrace$ respectively. Then $D\cong E$ .

Proof. [Sketch of Proof.] For every $i\in I$ , there are $j,k\in J$ , such that $\alpha(j)=\beta(k)$ . So $$E\to C_{\beta(k)}=E\to C_{\alpha(j)},$$ and its composition with $x_{\beta(k)}$ is the same as the composition with $x_{\alpha(j)}$ . By the universality of generalized pullbacks, we get a unique morphism $f:E\to D$ with $$E \stackrel{f}{\longrightarrow} D \longrightarrow C_{\beta(k)} = E \to C_{\beta(k)}.$$ Dually, we have a unique morphism $g:D\to E$ with $$D \stackrel{g}{\longrightarrow} E \longrightarrow C_{\alpha(j)} = D \to C_{\alpha(j)}.$$ As a result, $f\circ g=1_D$ and $g\circ f=1_E$ .