Extension of modules forms a set
I am reading the homological algebra by Hilton and Stammbach (Chapter 3.1) where an extension of modules $(A,B)$ is defined to be the exact sequences $0to Bto X to Ato0$ by an (well-understood) equivalent relation. We want to turn it into a set-valued bi-functor. But my question is: is it guaranteed that the quotient object is indeed a set? Can it be a proper class?
E.g. when we relax the restriction $0to Bto X$, then I do not think it is so true that all equivalent classes form a set.
commutative-algebra homological-algebra
asked Dec 12 '18 at 15:19
CO2CO2
1529
add a comment |
I am reading the homological algebra by Hilton and Stammbach (Chapter 3.1) where an extension of modules $(A,B)$ is defined to be the exact sequences $0to Bto X to Ato0$ by an (well-understood) equivalent relation. We want to turn it into a set-valued bi-functor. But my question is: is it guaranteed that the quotient object is indeed a set? Can it be a proper class?
E.g. when we relax the restriction $0to Bto X$, then I do not think it is so true that all equivalent classes form a set.
commutative-algebra homological-algebra
asked Dec 12 '18 at 15:19
CO2CO2
1529
$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28
add a comment |
I am reading the homological algebra by Hilton and Stammbach (Chapter 3.1) where an extension of modules $(A,B)$ is defined to be the exact sequences $0to Bto X to Ato0$ by an (well-understood) equivalent relation. We want to turn it into a set-valued bi-functor. But my question is: is it guaranteed that the quotient object is indeed a set? Can it be a proper class?
E.g. when we relax the restriction $0to Bto X$, then I do not think it is so true that all equivalent classes form a set.
commutative-algebra homological-algebra
asked Dec 12 '18 at 15:19
CO2CO2
1529
I am reading the homological algebra by Hilton and Stammbach (Chapter 3.1) where an extension of modules $(A,B)$ is defined to be the exact sequences $0to Bto X to Ato0$ by an (well-understood) equivalent relation. We want to turn it into a set-valued bi-functor. But my question is: is it guaranteed that the quotient object is indeed a set? Can it be a proper class?
E.g. when we relax the restriction $0to Bto X$, then I do not think it is so true that all equivalent classes form a set.
commutative-algebra homological-algebra
commutative-algebra homological-algebra
asked Dec 12 '18 at 15:19
CO2CO2
1529
asked Dec 12 '18 at 15:19
CO2CO2
1529
asked Dec 12 '18 at 15:19
CO2CO2
1529
asked Dec 12 '18 at 15:19
CO2CO2
1529
asked Dec 12 '18 at 15:19
CO2CO2
1529
1529
$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28
add a comment |
$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28
$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28
add a comment |
1 Answer
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Equivalence classes of extensions of $A$ by $B$ are in bijection with classes degree $1$ cocycles in a complex $hom(P,B)$ where $P$ is a projective resolution of $A$, so they are a set.
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
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Equivalence classes of extensions of $A$ by $B$ are in bijection with classes degree $1$ cocycles in a complex $hom(P,B)$ where $P$ is a projective resolution of $A$, so they are a set.
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
Equivalence classes of extensions of $A$ by $B$ are in bijection with classes degree $1$ cocycles in a complex $hom(P,B)$ where $P$ is a projective resolution of $A$, so they are a set.
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
add a comment |
Equivalence classes of extensions of $A$ by $B$ are in bijection with classes degree $1$ cocycles in a complex $hom(P,B)$ where $P$ is a projective resolution of $A$, so they are a set.
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
Equivalence classes of extensions of $A$ by $B$ are in bijection with classes degree $1$ cocycles in a complex $hom(P,B)$ where $P$ is a projective resolution of $A$, so they are a set.
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
answered Dec 12 '18 at 18:58
Pedro Tamaroff♦Pedro Tamaroff
96.4k10151296
96.4k10151296
add a comment |
add a comment |
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$X$ is equipotent to $Atimes B$.
– Pierre-Yves Gaillard
Dec 12 '18 at 23:17
@Pierre-YvesGaillard How to show this?
– CO2
Dec 13 '18 at 0:25
Each fiber of $Xto A$ is equipotent to $B$.
– Pierre-Yves Gaillard
Dec 13 '18 at 1:28