Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple category theory isomorphism

  1. May 8, 2009 #1
    Hey all,

    Okay, let me give this a wack. I want to show that [itex]A \times 1[/itex] is isomorphic to [itex]A[/itex]. I'm aware that this is trivial, even for a category theory style. However, sticking to the defs and conventions is tricky if you aren't aware of the subtleties, which is why I'm posting this. So here goes:

    Consider objects [itex]A[/itex] and [itex]A \times 1[/itex]. From the object [itex]A \times 1[/itex] we have the arrows [itex]\pi_1 : A \times 1 \to 1[/itex] and [itex]\pi_A : A \times 1 \to A[/itex]. Now we will also consider [itex]A[/itex] as a product in the following way: let [itex]\rho_1 : A \to 1[/itex] be the projection from A to 1, since this will always exists. Also, let [itex]\rho_{A \times 1} : A \to A \times 1[/itex] be the "projection" (really just a 'Cartesian inclusion'?) from A to [itex]A \times 1[/itex]

    To the expert: this last step I'm unsure about. It is obvious what it is from a set theory POV but from the category perspective it's not clear how the arrow might arise naturally (or legally).

    The rest of the proof is pretty straight forward: Since we have two products [itex]A[/itex] and [itex]A \times 1[/itex] we can compose [itex]\rho_{A \times 1} \circ \pi_A[/itex], which is a round trip on [itex]A \times 1[/itex], so it must be the identity. Similarly for [itex]\pi_A \circ \rho_{A \times 1}[/itex] must be the identity on A. And since these maps are unique and in opposite directions, they must be inverses, so we have a bijection between the two.

    - Farley
  2. jcsd
  3. May 9, 2009 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    You need to explain your hypotheses. What does x mean? Direct product - well that means you're not talking about a generic category? And what is 1? A category does not have an object that one calls '1' in general. Are you assuming an abelian category? Serre category? Grothendieck cateory?
  4. May 9, 2009 #3
    Sorry about that.. I shouldn't have tried to post late at night.

    Anyways, I'm just working with a vanilla category. However, as far as I've read, products are assumed, when they exist. And '1' is just a terminal object. So, instead, I would proceed:

    Let [itex]\mathcal{C}[/itex] be a category with [itex]1 \in Obj(\mathcal{C})[/itex] a terminal object and both [itex]A \times 1, A \in Obj(\mathcal({C}))[/itex] as products in this category. (Can you do that?)

    BTW this is proved on this guy's blog: http://unapologetic.wordpress.com/2007/06/27/categorification/ but he doesn't quite explain how f is both a projection and the unique map required by the product.. I thought I'd try a slightly different proof, hoping I didn't make a mistake.
  5. May 10, 2009 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    By the definition of direct product, there is a map A -> Ax1 as defined and it is unique (and it is not cartesian inclusion since there is no reason to suppose A and 1 are sets).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook