Talk:Quotient of an abelian category

Zero object

@AxelBoldt: I don’t think the statement of “treating the objects of the subcategory as zero objects” in the first paragraph of Quotient of an abelian category is very sound. For example, building on the vector space example given in the article, tensoring zero should result in zero (it’s not very common where one talks about monoidal abelian categories, but one certainly can, and one of the natural compatibility should be that $\operatorname {Hom} (A,C)\otimes _{\mathbb {Z} }\operatorname {Hom} (B,D)\rightarrow \operatorname {Hom} (A\otimes B,C\otimes D)$ is a homomorphism, but I’ve digressed), yet tensoring a finite dimensional vector space in the quotient category results in an isomorphic vector space, which in general isn’t a finite dimensional one, so clearly they aren’t treated as zero objects. I propose to delete this, leaving only the informal statement of “ignoring” the subcategory, since I do not know a better analogy. —Fourier-Deligne Transgirl (talk) 12:54, 7 November 2022 (UTC)

I don’t know if “wikiblaming” is a good practice here since I’m new to the community and haven’t had a lot of experience of contributing, but my reason is that I would prefer consulting the person who wrote this in the first place to see if I’ve missed anything. —Fourier-Deligne Transgirl (talk) 13:03, 7 November 2022 (UTC)

@Fourier-Deligne Transgirl: You should probably post your comments on their talk page, or on the talk page for the article you are discussing. This page should be used for discussing topics of broader interest. jraimbau (talk) 16:05, 7 November 2022 (UTC)

@Jean Raimbault: moved, thanks. --Fourier-Deligne Transgirl (talk) 18:32, 7 November 2022 (UTC)

I do think that "the objects of ${\mathcal {B}}$ become zero in the Serre quotient ${\mathcal {A/B}}$" is a good intuition to have. After all, the natural functor $Q:{\mathcal {A}}\to {\mathcal {A/B}}$ sends them to zero.

It is true (in the example discussed in the article) that $Q(X\otimes Y)\simeq Q(X)$ whenever $Y$ is finite-dimensional and $X$ is arbitrary. But this doesn't contradict the intuition "$Q(Y)$ is zero" since $Q(X\otimes Y)$ does not have to equal $Q(X)\otimes Q(Y)$. In other words: "$Y$ becomes 0" does not necessarily imply that "$X\otimes Y$ becomes $X\otimes 0$". I guess one could say that the Serre quotient is a construction for abelian categories and is not very well-behaved when applied to monoidal abelian categories. AxelBoldt (talk) 18:59, 7 November 2022 (UTC)