If there is an arrow $h: (A\times B) \to (A\times B)$, then, necessarily $h = \left\lt f,g\right\gt$ for some $f: (A\times B) \to A$ and $g (A\times B) \to B$?

The book I'm studying defines a product $A\times B$ to be an object provided with two projecting arrows $\pi_A : A\times B \to A$ and $\pi_B: A\times B \to B$ such that, for any other object C and arrows $f: C \to A$ and $g : C \to B$ there's exactly one arrow $\left\lt f,g\right\gt: C \to (A\times B)$ such that:

$\pi_A \cdot \left\lt f,g\right\gt = f$

$\pi_B \cdot \left\lt f,g\right\gt = g$

It says nothing about $\left\lt f,g\right\gt$ being the only arrow from $C$ to $A\times B$, but I got that impression later when he asked to prove that all products are isomorphic: I could only prove assuming that any arrow from $A \times B$ to itself was of the form $\left\lt f,g\right\gt$ for some f and g from $A\times B$ to A and B, respectively.

This suggests me that $\left\lt f,g\right\gt$ was meant to be the only possible arrow from C to $A \times B$.

Is this right or did I just misunderstood the book?

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    $\begingroup$ The answer to the question in your current first paragraph is: yes. Each map $f:A\times B\to A\times B$ is the product of the compositions $A\times B\stackrel{f}{\to} A\times B\stackrel{p_1}{\to}A$ and $A\times B\stackrel{f}{\to} A\times B\stackrel{p_2}{\to}B$, with $p_1:A\times B\to A$ and $p_2:A\times B\to B$ the projections. $\endgroup$ Jul 28, 2010 at 0:23
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    $\begingroup$ (The proof is left as an exercise for the reader!) $\endgroup$ Jul 28, 2010 at 0:24
  • $\begingroup$ Hummm! Now I see! If p1.<p1.f, p2.f> = p1.f and p2.<p1.f, p2.f> = p2.f and <p1.f,p2.f> must be the only arrow with this property, then necessarily f = <p1.f, p2.f>. Thanks. $\endgroup$
    – Rorsa
    Jul 28, 2010 at 0:38
  • $\begingroup$ Sorry for the confusing text, but the doubt was not very clear in my mind and I mixed two different things. $\endgroup$
    – Rorsa
    Jul 28, 2010 at 0:38

1 Answer 1


The reason you're having trouble proving the uniqueness is that you need to use the projections. There is not a unique map from $C$ to $A \times B$. But there is a unique map given specified maps from $C$ to $A$ and $B$ (i.e. such that the composition of the map from $C$ to the product with the projection gives the map from $C$ to $A$ or $B$). If $C_1$ and $C_2$ are two products, we can use the universal property to construct maps between them. The compositions are then the identity. Why? It is not true that there is only one map from $C_1$ to itself. BUT, there is only one map from $C_1$ to itself which, when composed with $\pi_A$ and $\pi_B$, gives $\pi_A$ and $\pi_B$. If you construct the maps correctly, this will be the case, and you will be able to prove the isomorphism between $C_1$ and $C_2$.


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