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Stream: learning: questions

Topic: Fin^op as a free object

James Deikun (May 07 2025 at 16:37):

$\mathbf{Fin}^\mathsf{op}$ is both the free FP category on one generator and the free finite limits category on one generator. Is there some reason this is the case or is it just a coincidence?

Mike Shulman (May 07 2025 at 16:41):

Similarly, for any discrete set $X$, the category $\mathrm{Set}^X$ is both the free coproduct-completion of $X$ and the free colimit-cocompletion of $X$. I would say that's because when $X$ is discrete, there are no interesting non-discrete diagrams in $X$ to take colimits of.

James Deikun (May 07 2025 at 16:45):

Ah, so even the non-discrete diagrams that appear in the completion end up reducing to discrete diagrams when you recast them over $X$? I guess then the interesting thing is that you even can recast them as diagrams over $X$.

Mike Shulman (May 07 2025 at 16:49):

Yes, it's not always the case that a free $\Phi$-cocompletion can be constructed in "one step" so that every object of the cocompletion is a $\Phi$-colimit of a diagram in the starting category, but it is the case for coproduct-completions and colimit-completions (and their finite versions).

Chaitanya Leena Subramaniam (May 08 2025 at 17:02):

It's worth noting that $\mathrm{FinSet}$ being both the finite coproduct-completion and the finite colimit-completion of $1$ (or for that matter, $\mathrm{Set}$ being both the free coproduct- and free colimit-completion of $1$) is only true in 1-category theory. In higher category theory ($(\infty,1)$-category theory or even $(n,1)$-category theory for $n>1$), $\mathrm{FinSet}$ is still the free finite coproduct-completion of $1$, but is no longer the free finite colimit-completion.

There are other 1-categories for which the free finite-colimit completion coincides with a completion under a strictly smaller class of diagrams. For instance, locally finite [[direct categories]] (direct categories whose slices are finite categories) have a finite-colimit completion that can be characterized as completion under finite cell complexes. This is a high-level explanation of why finite graphs or finite globular sets can be described as finite cell complexes.

Vít Jelínek (May 10 2025 at 22:31):

Chaitanya Leena Subramaniam said:

It's worth noting that $\mathrm{FinSet}$ being both the finite coproduct-completion and the finite colimit-completion of $1$ (or for that matter, $\mathrm{Set}$ being both the free coproduct- and free colimit-completion of $1$) is only true in 1-category theory. In higher category theory ($(\infty,1)$-category theory or even $(n,1)$-category theory for $n>1$), $\mathrm{FinSet}$ is still the free finite coproduct-completion of $1$, but is no longer the free finite colimit-completion.

The reason is that in higher categories, you have also copowers as finite colimits?
Or are you saying that FinSet is not even free cocompletion under conical colimits?

Mike Shulman (May 10 2025 at 22:40):

Not even conical colimits. In a higher setting, the coequalizer of $1\rightrightarrows 1$ is $S^1$.

James Deikun (May 10 2025 at 22:40):

You don't need copowers. Pullbacks and pushouts behave differently in higher categories. The colimit-completion of $1$ becomes $\mathrm{Fin}$ (finite spaces). As a result, the finite cocompletion is no longer "one-step" as described above.

James Deikun (May 10 2025 at 22:42):

Hm, or is it? Maybe if the coequalizer of two identity arrows is already a circle you don't need any diagrams that aren't made of identity arrows?

Mike Shulman (May 10 2025 at 22:59):

I think it's one-step, since any finite space is the realization of some finitely presented category (indeed, of a finite poset), so it should be the conical colimit of the constant diagram at 1 indexed by that category.

James Deikun (May 10 2025 at 23:08):

Ah, so that is how the [[Thomason model structure]] works!

Vít Jelínek (May 11 2025 at 00:38):

Oh, thanks!