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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