Product of objects in a category

Question

I have this definition. Let $A,B$ two objects in a category $\mathcal{C}$. A object $P$ in $\mathcal{C}$ is the product of $A$ and $B$ iff exist two morphisms $\phi: P \to A$ and $\psi: P \to B$ such that for all two morphisms $\varphi: A \to X$ and $\eta: B \to X$, exists an unique morphism $\pi: X \to P$ such that $\phi \pi = \varphi$ and $\psi \pi = \eta$. I have to prove that the product is determined except isomorphisms. But I don't know how to start the proof. Could anybody give an idea?

Answer 1

Let $\mathcal C$ be a category with products and let $X, Y$ be objects in $\mathcal C$. Then their product $X \times Y$ is unique up to isomorphism.

Proof: Suppose that there exists an object $P \neq X \times Y$ of $\mathcal C$ together with arrows $p_X : P \to X$ and $p_Y: P \to Y$ satisfying the universal property of the product of $X$ and $Y$. We can thus construct the following diagram in $\mathcal C$

where $m : P \to X \times Y$ is the unique arrow arising from the universal property of $P$ and $n : X \times Y \to P$ is the unique arrow arising from the universal property of $X \times Y$. Then, by the universal property of $X \times Y$, $n \circ m : X \times Y \to X \times Y$ is the unique arrow $X \times Y \to X \times Y$ making the resulting triangles commute. But $id_{X \times Y}$ also makes the resulting triangles commute, so $n \circ m = id_{X \times Y}$. An analogous argument shows that $m \circ n = id_{P}$.

Therefore, $m$ and $n$ are mutually inverse, so $P \cong X \times Y$, as required.