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
category-theory functors representable-functor hom-functor
edited Nov 22 at 10:49
user302797
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asked Nov 14 at 15:06
user122424
1,0681616
add a comment |
up vote
1
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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
category-theory functors representable-functor hom-functor
edited Nov 22 at 10:49
user302797
19.1k92251
asked Nov 14 at 15:06
user122424
1,0681616
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
Nov 14 at 18:53
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
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
Nov 14 at 22:05
add a comment |
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
category-theory functors representable-functor hom-functor
edited Nov 22 at 10:49
user302797
19.1k92251
asked Nov 14 at 15:06
user122424
1,0681616
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
category-theory functors representable-functor hom-functor
edited Nov 22 at 10:49
user302797
19.1k92251
asked Nov 14 at 15:06
user122424
1,0681616
edited Nov 22 at 10:49
user302797
19.1k92251
asked Nov 14 at 15:06
user122424
1,0681616
edited Nov 22 at 10:49
user302797
19.1k92251
edited Nov 22 at 10:49
user302797
19.1k92251
edited Nov 22 at 10:49
user302797
19.1k92251
19.1k92251
asked Nov 14 at 15:06
user122424
1,0681616
asked Nov 14 at 15:06
user122424
1,0681616
asked Nov 14 at 15:06
user122424
1,0681616
1,0681616
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
Nov 14 at 18:53
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
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
Nov 14 at 22:05
add a comment |
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
Nov 14 at 18:53
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
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
Nov 14 at 22:05
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
Nov 14 at 18:53
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
Nov 14 at 18:53
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
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
Nov 14 at 22:05
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
Nov 14 at 22:05
add a comment |
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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
Nov 14 at 18:53
I do not follow the meaning of the middle part of your sentence: " just with a richer diagram shape than the discrete one"
– user122424
Nov 14 at 21:07
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
Nov 14 at 22:05