In Hungerford's Algebra he defines a pullback of morphisms $f_1 \in \hom(X_1,A)$ and $f_2 \in \hom(X_2,A)$ as a commutative diagram $$\require{AMScd} \begin{CD} P @>{g_1}>>X_2\\ @V{g_2}VV @V{f_1}VV \\ X_2 @>{f_2}>> A \end{CD}$$ satisfying the universal property that for any commutative diagram $$\require{AMScd} \begin{CD} Q @>{h_1}>>X_2\\ @V{h_2}VV @V{f_1}VV \\ X_2 @>{f_2}>> A \end{CD}$$ there exists a unique morphism $t: Q \to P$ such that $h_i = g_i \circ t$. He then asks the reader to establish that
For any other pullback diagram with $P'$ in the upper-left corner $P \cong P'$.
How do we obtain this isomorphism?
The obvious choice seems to be considering the two morphisms $t: P \to P'$, $t': P' \to P$ and show that they compose to the identity. To this end,
$$h_1 = g_1 \circ t \implies h_1\circ 1 = h_1 \circ t' \circ t$$
but we cannot cancel unless $h_1$ is monic. Can we claim that necessarily $t \circ t'$ is the identity, since comparing $(P,g_1,g_2)$ with itself there exists a unique morphism $t'': P \to P$?