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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.

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

    matt grime

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    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

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    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).
     
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