Nearly locally presentable categories











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Here1, in the remark $2.3 (1)$ how from the fact that ${cal K}(A,-)$ does not preserve coproducts it follows that ${cal K}(A,-)$ sends special $lambda$-directed colimits to $lambda$-directed colimits and not to special $lambda$-directed ones?



1
Leonid Positselski, Jiří Rosický: Nearly locally presentable categories,
Theory and Appl. of Categories 33 (2018), #10, p.253-264;
http://www.tac.mta.ca/tac/volumes/33/10/33-10abs.html https://arxiv.org/abs/1710.10476










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  • To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
    – Kevin Carlson
    16 hours ago










  • I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
    – user122424
    14 hours ago










  • It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
    – Kevin Carlson
    13 hours ago















up vote
1
down vote

favorite












Here1, in the remark $2.3 (1)$ how from the fact that ${cal K}(A,-)$ does not preserve coproducts it follows that ${cal K}(A,-)$ sends special $lambda$-directed colimits to $lambda$-directed colimits and not to special $lambda$-directed ones?



1
Leonid Positselski, Jiří Rosický: Nearly locally presentable categories,
Theory and Appl. of Categories 33 (2018), #10, p.253-264;
http://www.tac.mta.ca/tac/volumes/33/10/33-10abs.html https://arxiv.org/abs/1710.10476










share|cite|improve this question
























  • To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
    – Kevin Carlson
    16 hours ago










  • I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
    – user122424
    14 hours ago










  • It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
    – Kevin Carlson
    13 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Here1, in the remark $2.3 (1)$ how from the fact that ${cal K}(A,-)$ does not preserve coproducts it follows that ${cal K}(A,-)$ sends special $lambda$-directed colimits to $lambda$-directed colimits and not to special $lambda$-directed ones?



1
Leonid Positselski, Jiří Rosický: Nearly locally presentable categories,
Theory and Appl. of Categories 33 (2018), #10, p.253-264;
http://www.tac.mta.ca/tac/volumes/33/10/33-10abs.html https://arxiv.org/abs/1710.10476










share|cite|improve this question















Here1, in the remark $2.3 (1)$ how from the fact that ${cal K}(A,-)$ does not preserve coproducts it follows that ${cal K}(A,-)$ sends special $lambda$-directed colimits to $lambda$-directed colimits and not to special $lambda$-directed ones?



1
Leonid Positselski, Jiří Rosický: Nearly locally presentable categories,
Theory and Appl. of Categories 33 (2018), #10, p.253-264;
http://www.tac.mta.ca/tac/volumes/33/10/33-10abs.html https://arxiv.org/abs/1710.10476







category-theory functors representable-functor hom-functor locally-presentable-categories






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edited 3 hours ago









Martin Sleziak

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user122424

1,0641616




1,0641616












  • To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
    – Kevin Carlson
    16 hours ago










  • I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
    – user122424
    14 hours ago










  • It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
    – Kevin Carlson
    13 hours ago


















  • To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
    – Kevin Carlson
    16 hours ago










  • I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
    – user122424
    14 hours ago










  • It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
    – Kevin Carlson
    13 hours ago
















To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
– Kevin Carlson
16 hours ago




To preserve special directed colimits, which are by definition always coproducts, just with a richer diagram shape than the discrete one, you would have to preserve coproducts.
– Kevin Carlson
16 hours ago












I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
14 hours ago




I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
14 hours ago












It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
– Kevin Carlson
13 hours ago




It's a coproduct expressed as a filtered colimit of sub-coproducts of up to size $lambda$ rather than as a discrete colimit. But it's still required to be a coproduct.
– Kevin Carlson
13 hours ago















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