Proving that pullback objects are unique up to isomorphism 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 $t'': P \to P$?
 A: As said by Martin Brandenburg in the comments, you stated the universal property wrong (not relevant anymore since the edit of the original post).
A pullback of $f_1\colon X_1 \to A,f_2\colon X_2 \to A$ is a diagram 
$$ \require{AMScd}
\begin{CD}
P @>{g_1}>>X_1\\
@V{g_2}VV @V{f_1}VV \\
X_2 @>{f_2}>> A
\end{CD} $$
satisfying that for any other diagram
$$\require{AMScd}
\begin{CD}
Q @>{h_1}>>X_1\\
@V{h_2}VV @V{f_1}VV \\
X_2 @>{f_2}>> A
\end{CD}$$
there exists a unique $t \colon Q \to P$ such that the following diagram commutes : 
             
             
             
             
    .

So now, if you have two pullback $P,P'$ there is $t \colon P \to P',t' \colon P' \to P$ such that commute the following diagrams :
             
             
             
             
    .
Notably, the arrow $t' \circ t \colon P \to P$ make the diagram 
             
             
             
             
   
commutes. By the universal property of the pullback $P$, such a $t'\circ t$ is unique : do you see another arrow $P \to P$ satisfying the same property ? Then it must equal $t' \circ t$.
Starting from here and elaborating a similar argument with the pullback $P'$, you should be able to prove the uniqueness up to isomorphism.
A: To answer the question at the very end of your post: yes. This sort of argument is fundamental, and applies in essentially the same way to any universal property (Wikipedia).
