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Here' what my book has to say on the universal property and specifically the universal property of the Cartesian product:
I understand the proof and that $\langle f,g\rangle$ is unique and all that. What I don't get is how this "property" supposedly characterizes the Cartesian product. The first paragraph says that universal properties "characterize a given construction" but don't we have to know what the Cartesian product is both to state the universal property and to prove it?
The meaning of universal property is in essence "the product is the smallest thing with two projections that behave in this way", in the sense that any other set $B$ with maps $B\to X$ and $B\to Y$ will factor uniquely through $X\times Y$. The product is an object $X\times Y$ for which the sets $\{\text{pair of maps } C\to X,C\to Y\}$ and the set $\{\text{maps }C\to X\times Y\}$ are in a certain natural bijection for every set $C$.
That being said, it is not unique! In any category, any objects determined by universal properties (limits, colimits..) are only unique up to isomorphism. For sets, isomorphism is the same as bijection and so any other set with equally many elements and a pair of maps in a similar way would also satisfy it.
Suppose we have some other thing $B$ that also satisfies the property: the universal property of such an object $B$ gives a unique map from every object with a pair of maps to $X$ and $Y$ and in particular, it gives a unique map $X\times Y\to B$. Obviously since $X\times Y$ also satisfies the property, we also have a unique map $B\to X\times Y$. By playing around with the properties, these unique maps can be shown to be isomorphisms, and thus we can be content with saying that $X\times Y$ is characterised "up to (some unique) isomorphism".
The Universal property characterizes the cartesian product the following way:
Theorem Let $Z$ be a set and $\pi^Z_1 : Z \to X$ and $\pi^Z_2:Z \to Y$ be any triple which satisfies the same universal property. Then, there exists some bijection $F : Z \to X \times Y$ such that for $j=1,2$ we have $$\pi_j \circ F = \pi_j^Z$$
Proof Since $\pi^Z_1 : Z \to X$ and $\pi^Z_2:Z \to Y$, by the Universal property from your Theorem there exists an unique $F : Z \to X \times Y$ such that $$\pi_j \circ F = \pi_j^Z$$
Now, since $\pi_1: X \times Y \to X$ and $\pi_2 : X \times Y \to Y$, by the universal property for $Z$ there exists some $G : X \times Y \to Z$ such that $$\pi^Z_j \circ G = \pi_j$$
Finally look at $F \circ G : X \times X \to X \times X$. Both $F \circ G$ and $Id$ satisfy the universal property for $<f,g>$ $$f=\pi_1 : X\ times Y \to X \,;\, g=\pi_2 : X \times Y \to Y$$ in your diagram. By the uniqeueness part we get $$F \circ G= Id$$
Same way, both $G \circ F$ and $Id$ satisfy the universal property for $Z$ hence $G \circ F =Id$.
This shows that $F,G$ are inverses of eachother.
