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Stream: learning: questions

Topic: ✔ categorical product from adjunction

simon (Aug 05 2025 at 18:12):

image.png

I meant $(Z \times Z, Z \times Z) \rightarrow (Z,Z)$ which is the component of $\epsilon$ at $(Z,Z)$, right? So this then creates the rectangle in diagram above and since it is the result of a natural transformation, it has to commute. Because of the existence of $h$ in $\mathcal{D}$, we have $(h,h)$ in $\mathcal{C}$, hence the arrow in my diagram. But does it necessarily form a commuting triangle with $(π_1,π_2)$ and $(p,q)$?
Sure, for a morphism $(Z \times Z,Z \times Z) \rightarrow (A,B)$ it makes sense to form a commuting triangle, since we compose $(Z \times Z,Z \times Z) \rightarrow (Z,Z)$ with $(p,q)$, two morphisms already forced to make the rectangle commute.

Ruby Khondaker (she/her) (Aug 05 2025 at 18:40):

Hm I'm getting a little confused here - the adjunction is $\text{Hom}(Z, A \times B) \cong \text{Hom}(\Delta(Z), (A, B))$. The counit comes from following the identity along $\text{Hom}(A \times B, A \times B) \cong \text{Hom}(\Delta(A \times B), (A, B))$ which gets you the projections, and the unit comes from following the identity along $\text{Hom}(Z, Z \times Z) \cong \text{Hom}(\Delta(Z), \Delta(Z))$. It's not entirely clear to me where this morphism $\Delta(Z \times Z) \to \Delta(Z)$ comes from?

I guess you could follow the identity along $\text{Hom}(Z \times Z, Z \times Z) \cong \text{Hom}(\Delta(Z \times Z), (Z, Z))$, which gets you the projections onto the first and second factor?

simon (Aug 05 2025 at 19:07):

I guess you could follow the identity along $\text{Hom}(Z \times Z, Z \times Z) \cong \text{Hom}(\Delta(Z \times Z), (Z, Z))$, which gets you the projections onto the first and second factor?

Exactly. So I try to see if I can get to the commuting triangle we need for the product by covering up the hom-set definition, just looking at both categories and the unit and co-unit. But I guess that is too weak as the universal morphism definition mentions "... exists a unique $h$ such that $(π_1,π_2) \circ (h,h) = (p,q)$ (i.e. commutes)". That is why I was talking about the commuting rectangle because I thought, hmm maybe we can squeeze out the commuting triangle out of it that we need for the commuting triangle that falls in place with the universal property definition of the product. I mean, you can see that seeing it through the lens of the universal property definition is most likely not doing me a favour here.

Ruby Khondaker (she/her) (Aug 05 2025 at 19:13):

So there is a general way to go from a homset characterisation to a unit-counit characterisation; what you'd be looking for are the "triangular identities".

simon (Aug 05 2025 at 19:19):

Yeah, I already saw them as well. I think I'll allow myself to let it sink in and probably come back later if any other questions come up. When I started looking at videos and comments, it was already clear to me that adjunctions are not the most accessible topic in category theory. Thank you for answering my questions and for you patience. It is much appreciated!

Notification Bot (Aug 05 2025 at 19:19):

simon has marked this topic as resolved.