Prove the existence of one and only one isomorphism between products of objects of a category

Question

Let $ \mathscr{C} $ be a category, $ X_1 $ and $ X_2 $ two objects of $ \mathscr{C} $ and $ (P,p_1,p_2) $ and $ (Q,q_1,q_2) $ two products of $ X_1 $ and $ X_2 $. Prove that there is one and only one isomorphism $ \psi:Q\to P $ such that $ p_1\circ \psi=q_1 $ and $ p_2\circ \psi=q_2 $.

Here is my attempt to answer:

Since $ Q $ is an object in $ \mathscr{C} $, there exists a unique morphism $ \psi:Q\to P=X_1\times X_2 $ such that $ p_1\circ \psi=q_1 $ and $ p_2\circ \psi=q_2 $ by the definition of products. Similarly, there exists a unique morphism $ \psi':P\to Q=X_1\times X_2 $ such that $ q_1\circ \psi'=p_1 $ and $ q_2\circ \psi'=p_2 $. Hence $$ q_1=p_1\circ \psi=(q_1\circ \psi')\circ \psi, $$ $$ q_2=p_2\circ \psi=(q_2\circ \psi')\circ \psi, $$ $$ p_1=q_1\circ \psi'=(p_1\circ \psi)\circ \psi' $$ and $$ p_2=q_2\circ \psi'=(p_2\circ \psi)\circ \psi' $$ Can I conclude that $ \psi'=\psi^{-1} $, the inverse of $ \psi $ and that $ \psi $ therefore is an isomorphism?

Answer 1

Your almost there! You found a good candidate $\psi$, and a candidate for its inverse $\psi'$. So all you have to show is that $\psi'\psi = \mathrm{id}$ and $\psi\psi' = \mathrm{id}$.

Now, rewriting a little bit your equations through associativity of composition, you also proved that $f = \psi'\psi$ is an arrow $Q \to Q$ satisfying: $$ q_1 = q_1 f \quad \text{and} \quad q_2 = q_2 f $$ Do you know such another $f$? If so, what does it tell you on $f$? (Hint: use the universal property of the product $Q$.)

Symmetrically you proved that $g = \psi\psi'$ is an arrow $P \to P$ satisfying: $$ p_1 = p_1 g \quad \text{and} \quad p_2 = p_2 g $$ Play the same game on $g$ with the universal property of the product $P$.