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By bicolimit I mean what Kelly means in its "Elementary observations on 2-categorical limits". If we have a diagram (pseudofunctor) $G\colon\mathcal P\to\mathcal K$, the bicolimit of $G$ is the (unique up to equivalence) object $\mathsf{bicolim}G$ in $\mathcal K$ such that for every object $A$ we have a natural equivalence of categories

$$\mathcal K(\mathsf{bicolim}G,A)\simeq\mathsf{PsNat}(\Delta\ast,\mathcal K(G-,A)),$$

where $\mathsf{PsNat}$ is the category having modifications as 1-cells, and $\Delta\ast$ is the constant pseudofunctor $\mathcal P^\mathsf{op}\to\mathsf{Cat}$. Now my aim is to prove that for $\mathcal K=\mathsf{Add}$, the bicategory of additive categories, additive functors and natural transformations is indeed bicocomplete. Kelly proves this for $\mathsf{Cat}$ with an argument that leads to something more, that is an isomorphism of the above categories. This fact, and the feeling that I suppose the equivalence not to be necessary an isomorphism for $\mathsf{Add}$, suggests me that we cannot use the same argument in this context.

Are there explicit constructions of a bicolimit for categories that we can mimick in the additive case? Or, even better, are there some general results that may turn applicable to this setting?

Thank you so much!

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    $\begingroup$ I am pretty sure that you can build them exactly as for regular categories - I have explained this case in arxiv.org/abs/2001.10123 which also contains some general (well-known) construction principles: it suffices to build coproducts, coinserters and coequifiers. $\endgroup$ Commented Sep 25, 2023 at 16:20

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Yes, additive categories are $Ab$-enriched categories with all finite direct sums. $Ab$-enriched categories have all 2-limits and 2-colimits by enriched category theory and in particular all bilimits and bicolimits. Additive categories are bireflective inside $Ab$-enriched categories, so they have bilimits and bicolimits as well.

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    $\begingroup$ It would be nice to give some references for these facts. $\endgroup$
    – varkor
    Commented Sep 23, 2023 at 7:40
  • $\begingroup$ Thank you so much for your answer. I would be glad indeed to consult some references too :) $\endgroup$
    – Nikio
    Commented Sep 24, 2023 at 20:30

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