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Stream: learning: questions

Topic: Should we replace two objects with their product?

Julius Hamilton (Sep 21 2024 at 18:42):

If it turns out that for two fixed objects $X, Y \in Ob(C)$, $X \times Y$ exists, does that mean we should just “forget about $X$ and $Y$”? It sounds like we can make our category simpler by getting rid of $X$ and $Y$.

John Baez (Sep 21 2024 at 20:33):

No, you should not forget about $X$ and $Y$. For example, the 6-element set $\{1, 2, 3, 4, 5, 6 \}$ is the product of $\{1, 2\}$ and $\{1,2,3\}$. But this does not mean we should forget about those other sets.

All this is a fancy way of saying "just because $6 = 2 \times 3$, that doesn't mean you should get rid of $2$ and $3$". For one thing, if you did, you couldn't say $6 = 2 \times 3$ anymore! For another, I really like the numbers $2$ and $3$.

Julius Hamilton (Sep 21 2024 at 22:40):

Awesome, thx

Madeleine Birchfield (Sep 22 2024 at 03:52):

Part of the definition of a category theoretic product $X \times Y$ is the data of $X$ and $Y$, since one needs them to define the morphisms from $X \times Y$ to $X$ and $Y$ and then to assert the universal property that this span is the terminal span to $X$ and $Y$.

David Michael Roberts (Sep 22 2024 at 08:38):

I think John's example doesn't convey the sense of violence of removing sets from the category. If one removed _all_ two-element sets and all three-element sets (which is the only thing that would make a material difference, not just removing one such pair), then your category suddenly is much poorer for structure. Certain colimits would fail to exist, I think. Moreover, if you did this with one-element sets, since every set is isomorphic to the product of itself with a one-element set, you would lose terminal objects, and in fact you could accidentally throw out all the objects!

Julius Hamilton (Sep 23 2024 at 13:57):

I see, thank you.