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Stream: learning: questions

Topic: Final "twisted diagonal" functors

fosco (Sep 11 2020 at 09:38):

It is well known that given a category $\cal C$ the following two conditions are equivalent:

1. the diagonal functor $\Delta : {\cal C} \to {\cal C}\times {\cal C}$ is final
2. For every two objects $X,Y$ of ${\cal C}$, the category of cospans $X \to A\leftarrow Y$ is (nonempty and?) connected.

We say that ${\cal C}$ is sifted if any of those conditions is true. It takes very little effort to verify that if ${\cal C}$ is sifted, then every diagonal functor ${\cal C} \to {\cal C}^n$ to a finite power of ${\cal C}$ is also final.

Where I can look for an explicit proof of the above equivalence? It seems to me that it entails the following: let $\nabla$ be the "twisted diagonal functor" $\nabla : {\cal C}^\text{op}\times {\cal C} \to {\cal C}^\text{op}\times {\cal C}^\text{op}\times {\cal C} \times {\cal C}$, defined as $\Delta_{{\cal C}^\text{op}}\times \Delta_{\cal C}$. If ${\cal C}$ is sifted and cosifted, for example if it has products and coproducts, then $\nabla$ is final.

Is this correct, or I'm losing something?

Nathanael Arkor (Sep 11 2020 at 10:42):

Which equivalence are you looking for? Aren't (1) and (2) the same, by unwrapping the definition of "final"?

Nathanael Arkor (Sep 11 2020 at 10:44):

("Connected" implies nonempty.)

fosco (Sep 11 2020 at 12:22):

In hindsight that was a naive question... I agree there's almost nothing to prove for 1 iff 2.

I'm looking for properties of $\cal C$ that imply $\nabla$ is final.

Morgan Rogers (he/him) (Sep 11 2020 at 12:38):

If you rearrange the components in the codomain of $\nabla$, how does it compare to the diagonal functor for $\mathcal{C}^{\mathrm{op}} \times \mathcal{C}$?

Morgan Rogers (he/him) (Sep 11 2020 at 12:38):

If it's the same, you have your answer as a condition on that product category