most recent 30 from http://mathoverflow.net 2013-05-21T09:47:26Z http://mathoverflow.net/feeds/question/33591 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33591/products-of-objects-in-categories Rorsa 2010-07-28T00:11:28Z 2010-07-28T00:27:44Z

<p>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 $g (A\times B) \to B$?</p> <p>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:</p> <p>$\pi_A \cdot \left\lt f,g\right\gt = f$</p> <p>$\pi_B \cdot \left\lt f,g\right\gt = g$</p> <p>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.</p> <p>This suggests me that $\left\lt f,g\right\gt$ was meant to be the only possible arrow from C to $A \times B$.</p> <p>Is this right or did I just misunderstood the book?</p>

http://mathoverflow.net/questions/33591/products-of-objects-in-categories/33593#33593 David Corwin 2010-07-28T00:27:44Z 2010-07-28T00:27:44Z

<p>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 <em>given</em> 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 $C_2$.</p>