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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeSep 11th 2012

    touched the formatting at additive category

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeMay 18th 2016

    I have added in the detailed proof of the proposition (here) that in an Ab-enriched category all finite (co-)products are biproducts.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 18th 2016

    For completeness, further below in the Properties-section (starting here) I have spelled out the way semiadditive structure induces enrichment in commutative monoids, and that this induced enrichment coincides with the original enrichement if we started with an additive category.

    These statements are scattered over other entries already, of course, but for readability if may be good to have them here in one place.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2016

    I have expanded just a little more the (elementary) proof that in an Ab-enriched category finite products are biproducts (here). Maybe somewhat pedantically, but just to be completely clear.

    • CommentRowNumber5.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 7th 2016

    Note that subtraction is not needed: the result holds for CMonCMon-enriched categories. In fact, having biproducts implies CMonCMon-enrichment.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2016
    • (edited Jun 7th 2016)

    Yes, that’s discussed at biproduct. But since I am editing the entry on additive categories, I am talking there about Ab-enrichment.

    [actually it’s also discussed further below in the entry on additive categories]

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeJun 7th 2016

    I have also made more explicit the (elementary) proofs of this prop. and this prop.

    • CommentRowNumber8.
    • CommentAuthornaughie
    • CommentTimeJul 24th 2019

    In the proof of the proposition that finite products coincide with finite coproducts in a Ab-enriched category, the zero-ary case holds if the category has both an initial object and a terminal object. But, the existence of a terminal object (assuming the existence of an initial object) is not so clear to me.

    • CommentRowNumber9.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 24th 2019

    Well, if 00 is initial, then for any object AA the zero morphism z:A0z: A \to 0 is available and, by initiality, is left inverse to the unique map !:0A!: 0 \to A which is also the zero element in hom(0,A)\hom(0, A). So !! is monic; given any f:A0f: A \to 0, the composites !f! f and !z! z must both be the zero element in hom(A,A)\hom(A, A) since composition preserves zero morphisms (by enrichment). Then f=zf = z by monicity, so there is exactly one map z:A0z: A \to 0.

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeJul 24th 2019

    However, the proof as currently given on the page shows less than the proposition claims. The proposition claims that “any terminal object is also an initial object”, but the proof given on the page (in contrast to the one Todd just gave) only shows (as naughie said) that if an additive category has both an initial and terminal object then they are isomorphic. Similarly, the proposition claims that any finite product is also a coproduct, but the proof given for binary products assumes the existence of a zero object.

    • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 24th 2019

    Strengthened the argument that in CMonCMon-enriched categories, terminal objects are initial and conversely.

    diff, v34, current

    • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeJul 24th 2019

    Thanks. I guess I wasn’t looking hard enough in the other case – the proof for binary products uses only zero morphisms, not zero objects. So I think it’s all good now.

    • CommentRowNumber13.
    • CommentAuthornaughie
    • CommentTimeJul 30th 2019

    Thanks, but I found a simpler proof: if AA is an initial object, then id A=0\mathrm{id}_A = 0 because it is the only morphism AAA \to A. So, for any morphism f:BAf: B \to A, we have f=id Af=0f=0f = \mathrm{id}_A f = 0 f = 0; hence AA should be terminal.

    • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 30th 2019

    Congratulations, naughie. I’ll be glad to edit that in later.

    • CommentRowNumber15.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 30th 2019

    Implemented naughie’s improvement.

    diff, v36, current

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