Mixing
When does $mathrmExt^ast(K,-)$ preserve filtered colimits?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Fix an abelian category $mathcalA$ that admits filtered colimits, and for $K,HinmathcalA$, write $mathcalA^n(K,H) = mathrmExt^n(K,H)$ for the group of Yoneda $n$-extensions of $K$ by $H$, definable in any abelian category. My question is:
- Under what conditions on $mathcalA$ and $K$ does $mathcalA^n(K,-)colon mathcalArightarrowmathrmSet$ preserve filtered colimits?
I would be happy with an answer for $n=1$. In the case $n=0$, an object $K$ such that $mathcalA(K,-)$ preserves filtered colimits is called compact. However, it does not appear that this is sufficient for $mathcalA^n(K,-)$ to preserve filtered colimits, as the former is concerned with maps out of $K$ and the latter with maps into $K$.
I can prove that $mathcalA^1(K,-)$ does preserve filtered colimits in the somewhat contrived situation where $mathcalA$ is a compactly-generated abelian category and $KinmathcalA$ a compact object such that:
If $GinmathcalA$ is compact, then every subobject of $G$ is compact,
If $PsubsetmathrmSub(K)$ is a filtered subposet such that $bigvee P = K$, then $Kin P$.
The first condition fails in case $mathcalA = mathrmMod_R$ and $K=R$ for a ring $R$ which is not Noetherian, so these conditions are rather suboptimal. I do not know an example where the second condition fails.
-
Here is a sketch of the proof under these conditions. We must show $mathrmcolim_j mathcalA^1(K,H_j)rightarrowmathcalA^1(K,H)$ is an isomorphism for all filtered colimits $H=mathrmcolim_j H$.
For surjectivity, it is sufficient by compact generation to show that every element $xin mathcalA^1(K,H)$ can be written as $x=f_ast(y)$, where $fcolon H'rightarrow H$ with $H'$ compact and $yinmathcalA^1(K,H')$. Given an extension $0rightarrow Hrightarrow Grightarrow Krightarrow 0$, we may write $G=mathrmcolim_jinmathcalJ G_j$ with $mathcalJ$ filtered and $G_j$ compact, and by condition (ii) there is some $j$ such that $G_jrightarrow K$ is epi. If $H' = Htimes_G G_j$, then $H'$ is compact by condition (i), and the extension is the image of $0rightarrow H'rightarrow G_jrightarrow Krightarrow 0$.
For injectivity, say $H=mathrmcolim_j H_j$ as a filtered colimit. The sequences $0rightarrow H_jrightarrow Hrightarrow H/H_jrightarrow 0$ give an exact sequence $mathrmcolim_j mathcalA(K,H/H_j)rightarrowmathrmcolim_jmathcalA^1(K,H_j)rightarrowmathrmcolim_jmathcalA^1(K,H)$, and the first term is zero as $K$ is compact and $mathrmcolim_j H/H_j = 0$.
category-theory homological-algebra abelian-categories
asked Jul 17 at 17:24
ne-
1067
add a comment |Â
up vote
2
down vote
favorite
Fix an abelian category $mathcalA$ that admits filtered colimits, and for $K,HinmathcalA$, write $mathcalA^n(K,H) = mathrmExt^n(K,H)$ for the group of Yoneda $n$-extensions of $K$ by $H$, definable in any abelian category. My question is:
- Under what conditions on $mathcalA$ and $K$ does $mathcalA^n(K,-)colon mathcalArightarrowmathrmSet$ preserve filtered colimits?
I would be happy with an answer for $n=1$. In the case $n=0$, an object $K$ such that $mathcalA(K,-)$ preserves filtered colimits is called compact. However, it does not appear that this is sufficient for $mathcalA^n(K,-)$ to preserve filtered colimits, as the former is concerned with maps out of $K$ and the latter with maps into $K$.
I can prove that $mathcalA^1(K,-)$ does preserve filtered colimits in the somewhat contrived situation where $mathcalA$ is a compactly-generated abelian category and $KinmathcalA$ a compact object such that:
If $GinmathcalA$ is compact, then every subobject of $G$ is compact,
If $PsubsetmathrmSub(K)$ is a filtered subposet such that $bigvee P = K$, then $Kin P$.
The first condition fails in case $mathcalA = mathrmMod_R$ and $K=R$ for a ring $R$ which is not Noetherian, so these conditions are rather suboptimal. I do not know an example where the second condition fails.
-
Here is a sketch of the proof under these conditions. We must show $mathrmcolim_j mathcalA^1(K,H_j)rightarrowmathcalA^1(K,H)$ is an isomorphism for all filtered colimits $H=mathrmcolim_j H$.
For surjectivity, it is sufficient by compact generation to show that every element $xin mathcalA^1(K,H)$ can be written as $x=f_ast(y)$, where $fcolon H'rightarrow H$ with $H'$ compact and $yinmathcalA^1(K,H')$. Given an extension $0rightarrow Hrightarrow Grightarrow Krightarrow 0$, we may write $G=mathrmcolim_jinmathcalJ G_j$ with $mathcalJ$ filtered and $G_j$ compact, and by condition (ii) there is some $j$ such that $G_jrightarrow K$ is epi. If $H' = Htimes_G G_j$, then $H'$ is compact by condition (i), and the extension is the image of $0rightarrow H'rightarrow G_jrightarrow Krightarrow 0$.
For injectivity, say $H=mathrmcolim_j H_j$ as a filtered colimit. The sequences $0rightarrow H_jrightarrow Hrightarrow H/H_jrightarrow 0$ give an exact sequence $mathrmcolim_j mathcalA(K,H/H_j)rightarrowmathrmcolim_jmathcalA^1(K,H_j)rightarrowmathrmcolim_jmathcalA^1(K,H)$, and the first term is zero as $K$ is compact and $mathrmcolim_j H/H_j = 0$.
category-theory homological-algebra abelian-categories
asked Jul 17 at 17:24
ne-
1067
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Fix an abelian category $mathcalA$ that admits filtered colimits, and for $K,HinmathcalA$, write $mathcalA^n(K,H) = mathrmExt^n(K,H)$ for the group of Yoneda $n$-extensions of $K$ by $H$, definable in any abelian category. My question is:
- Under what conditions on $mathcalA$ and $K$ does $mathcalA^n(K,-)colon mathcalArightarrowmathrmSet$ preserve filtered colimits?
I would be happy with an answer for $n=1$. In the case $n=0$, an object $K$ such that $mathcalA(K,-)$ preserves filtered colimits is called compact. However, it does not appear that this is sufficient for $mathcalA^n(K,-)$ to preserve filtered colimits, as the former is concerned with maps out of $K$ and the latter with maps into $K$.
I can prove that $mathcalA^1(K,-)$ does preserve filtered colimits in the somewhat contrived situation where $mathcalA$ is a compactly-generated abelian category and $KinmathcalA$ a compact object such that:
If $GinmathcalA$ is compact, then every subobject of $G$ is compact,
If $PsubsetmathrmSub(K)$ is a filtered subposet such that $bigvee P = K$, then $Kin P$.
The first condition fails in case $mathcalA = mathrmMod_R$ and $K=R$ for a ring $R$ which is not Noetherian, so these conditions are rather suboptimal. I do not know an example where the second condition fails.
-
Here is a sketch of the proof under these conditions. We must show $mathrmcolim_j mathcalA^1(K,H_j)rightarrowmathcalA^1(K,H)$ is an isomorphism for all filtered colimits $H=mathrmcolim_j H$.
For surjectivity, it is sufficient by compact generation to show that every element $xin mathcalA^1(K,H)$ can be written as $x=f_ast(y)$, where $fcolon H'rightarrow H$ with $H'$ compact and $yinmathcalA^1(K,H')$. Given an extension $0rightarrow Hrightarrow Grightarrow Krightarrow 0$, we may write $G=mathrmcolim_jinmathcalJ G_j$ with $mathcalJ$ filtered and $G_j$ compact, and by condition (ii) there is some $j$ such that $G_jrightarrow K$ is epi. If $H' = Htimes_G G_j$, then $H'$ is compact by condition (i), and the extension is the image of $0rightarrow H'rightarrow G_jrightarrow Krightarrow 0$.
For injectivity, say $H=mathrmcolim_j H_j$ as a filtered colimit. The sequences $0rightarrow H_jrightarrow Hrightarrow H/H_jrightarrow 0$ give an exact sequence $mathrmcolim_j mathcalA(K,H/H_j)rightarrowmathrmcolim_jmathcalA^1(K,H_j)rightarrowmathrmcolim_jmathcalA^1(K,H)$, and the first term is zero as $K$ is compact and $mathrmcolim_j H/H_j = 0$.
category-theory homological-algebra abelian-categories
asked Jul 17 at 17:24
ne-
1067
Fix an abelian category $mathcalA$ that admits filtered colimits, and for $K,HinmathcalA$, write $mathcalA^n(K,H) = mathrmExt^n(K,H)$ for the group of Yoneda $n$-extensions of $K$ by $H$, definable in any abelian category. My question is:
- Under what conditions on $mathcalA$ and $K$ does $mathcalA^n(K,-)colon mathcalArightarrowmathrmSet$ preserve filtered colimits?
I would be happy with an answer for $n=1$. In the case $n=0$, an object $K$ such that $mathcalA(K,-)$ preserves filtered colimits is called compact. However, it does not appear that this is sufficient for $mathcalA^n(K,-)$ to preserve filtered colimits, as the former is concerned with maps out of $K$ and the latter with maps into $K$.
I can prove that $mathcalA^1(K,-)$ does preserve filtered colimits in the somewhat contrived situation where $mathcalA$ is a compactly-generated abelian category and $KinmathcalA$ a compact object such that:
If $GinmathcalA$ is compact, then every subobject of $G$ is compact,
If $PsubsetmathrmSub(K)$ is a filtered subposet such that $bigvee P = K$, then $Kin P$.
The first condition fails in case $mathcalA = mathrmMod_R$ and $K=R$ for a ring $R$ which is not Noetherian, so these conditions are rather suboptimal. I do not know an example where the second condition fails.
-
Here is a sketch of the proof under these conditions. We must show $mathrmcolim_j mathcalA^1(K,H_j)rightarrowmathcalA^1(K,H)$ is an isomorphism for all filtered colimits $H=mathrmcolim_j H$.
For surjectivity, it is sufficient by compact generation to show that every element $xin mathcalA^1(K,H)$ can be written as $x=f_ast(y)$, where $fcolon H'rightarrow H$ with $H'$ compact and $yinmathcalA^1(K,H')$. Given an extension $0rightarrow Hrightarrow Grightarrow Krightarrow 0$, we may write $G=mathrmcolim_jinmathcalJ G_j$ with $mathcalJ$ filtered and $G_j$ compact, and by condition (ii) there is some $j$ such that $G_jrightarrow K$ is epi. If $H' = Htimes_G G_j$, then $H'$ is compact by condition (i), and the extension is the image of $0rightarrow H'rightarrow G_jrightarrow Krightarrow 0$.
For injectivity, say $H=mathrmcolim_j H_j$ as a filtered colimit. The sequences $0rightarrow H_jrightarrow Hrightarrow H/H_jrightarrow 0$ give an exact sequence $mathrmcolim_j mathcalA(K,H/H_j)rightarrowmathrmcolim_jmathcalA^1(K,H_j)rightarrowmathrmcolim_jmathcalA^1(K,H)$, and the first term is zero as $K$ is compact and $mathrmcolim_j H/H_j = 0$.
category-theory homological-algebra abelian-categories
asked Jul 17 at 17:24
ne-
1067
edited Jul 17 at 17:36
asked Jul 17 at 17:24
ne-
1067
asked Jul 17 at 17:24
ne-
1067
asked Jul 17 at 17:24
ne-
1067
1067
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(prod_i y_i,x)cong prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(prod_i y_i,x)cong prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
add a comment |Â
up vote
1
down vote
accepted
If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(prod_i y_i,x)cong prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(prod_i y_i,x)cong prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
If your abelian category is a module category, then this is well understood. To get arbitrary exts to preserve filtered colimits, you want your module to be quasi-isomorphic to a complex of finitely generated projectives. Brown proved this in his paper "A homological criterion for finiteness." Over a Noetherian ring, this is just asking for your module to be finitely generated, which generalizes the special case of your result that is he case of modules over a Noetherian ring.
The assumptions in your argument aren't too insane: that subobjects of compacts be compact is an abstract Noetherian condition, while the condition on unions of filtered families of subobjects always holds. Indeed, since $K$ is compact, its identity morphism factors through some subobject in the family! However, an abelian category in which every object is a filtered colimit of compacts is always a reflective subcategory, close down under filtered colimits, of the category of modules over a "ring with many objects", that is, a small preadditive category. So your version and Brown's have surprisingly close to the same scope, and I'm sure Brown's theorem generalizes to your setting.
In more general abelian categories, there's no hope, because in general there are not even any objects whose non-derived homs preserve filtered colimits. For instance, such an object x of the opposite of a module category would satisfy, in the module category, $Hom(prod_i y_i,x)cong prod_i Hom(y_i,x).$ This is absurd, as follows for $x, y_i$ all the same finite cyclic group of prime order by a dimension count and for $x, y_i$ all the integers by Specker's theorem that the dual of a countable direct product of copies of the integers is the countable direct sum.
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
edited Jul 17 at 19:34
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
answered Jul 17 at 19:28
Kevin Carlson
29.2k23065
29.2k23065
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
add a comment |Â
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
Thank you for your answer, I think this clears up some things in my mind. I certainly agree that some kind of accessibility condition is necessary for anything general to be said. My Noetherian condition still bothers me. If I recall, in $R$-modules, $M$ is compact iff $M$ is f.p.; so I guess what would make me happy is a categorical formulation of "$n$-f.p." (having a resolution with first $n+1$ terms f.g.) that would work in general without choosing a ringoid $R$. But I don't really expect a good formulation to exist.
– ne-
Jul 17 at 20:07
1
1
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
By the way, the paper by Brown you reference appears to be titled "Homological Criteria for Finiteness". This might sound pointlessly picky, but "A Homological Criteria for Finiteness" is a distinct (related) paper by Strebel.
– ne-
Jul 17 at 20:11
1
1
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
(If I think about it from the other direction, I guess that asking for $mathcalA^n(K,-)$ to preserve filtered colimits already is a good categorical formulation of being "$n+1$-f.p.").
– ne-
Jul 17 at 20:39
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854735%2fwhen-does-mathrmext-astk-preserve-filtered-colimits%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password