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Connecting homomorphism anti-commutative.
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This is a part of statement made in Cartan Eilenberg Homological Algebra Pg 44 on Satellite.
Given $0to A^0to A^1tocdotsto A^qto 0$, one can induce connecting homomorphism by short exact sequence by applying satellite construction as satellite functor induces long exact sequence. Say $T=T^n$ is a family of additive functor with connecting homomorphism $T^n(A'')to T^n+1(A')$ for any short exact sequence $0to A'to Ato A''to 0$. Consider the following commutative diagram.
$0to A'to Ato A''to 0$
$0to B'to Bto B''to 0$
$0to C'to Cto C''to 0$
There will be arrows in vertical direction s.t above diagram is commutative and exact in columns.
Then there is an anticommutative diagram resulted. The morphisms $T^n-1(C'')to T^n(C')to T^n+1(A')$ and $T^n-1(C'')to T^n(A'')to T^n+1(A')$ differ by a minus sign.
$textbfQ:$ Where is the origin of this anticommutative diagram? Any particular reason to expect this? I am aware this happens in homology part. Is this hinting the bicomplex for spectral sequence?
abstract-algebra homological-algebra
asked Jul 17 at 23:01
user45765
2,1942718
add a comment |Â
up vote
2
down vote
favorite
This is a part of statement made in Cartan Eilenberg Homological Algebra Pg 44 on Satellite.
Given $0to A^0to A^1tocdotsto A^qto 0$, one can induce connecting homomorphism by short exact sequence by applying satellite construction as satellite functor induces long exact sequence. Say $T=T^n$ is a family of additive functor with connecting homomorphism $T^n(A'')to T^n+1(A')$ for any short exact sequence $0to A'to Ato A''to 0$. Consider the following commutative diagram.
$0to A'to Ato A''to 0$
$0to B'to Bto B''to 0$
$0to C'to Cto C''to 0$
There will be arrows in vertical direction s.t above diagram is commutative and exact in columns.
Then there is an anticommutative diagram resulted. The morphisms $T^n-1(C'')to T^n(C')to T^n+1(A')$ and $T^n-1(C'')to T^n(A'')to T^n+1(A')$ differ by a minus sign.
$textbfQ:$ Where is the origin of this anticommutative diagram? Any particular reason to expect this? I am aware this happens in homology part. Is this hinting the bicomplex for spectral sequence?
abstract-algebra homological-algebra
asked Jul 17 at 23:01
user45765
2,1942718
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is a part of statement made in Cartan Eilenberg Homological Algebra Pg 44 on Satellite.
Given $0to A^0to A^1tocdotsto A^qto 0$, one can induce connecting homomorphism by short exact sequence by applying satellite construction as satellite functor induces long exact sequence. Say $T=T^n$ is a family of additive functor with connecting homomorphism $T^n(A'')to T^n+1(A')$ for any short exact sequence $0to A'to Ato A''to 0$. Consider the following commutative diagram.
$0to A'to Ato A''to 0$
$0to B'to Bto B''to 0$
$0to C'to Cto C''to 0$
There will be arrows in vertical direction s.t above diagram is commutative and exact in columns.
Then there is an anticommutative diagram resulted. The morphisms $T^n-1(C'')to T^n(C')to T^n+1(A')$ and $T^n-1(C'')to T^n(A'')to T^n+1(A')$ differ by a minus sign.
$textbfQ:$ Where is the origin of this anticommutative diagram? Any particular reason to expect this? I am aware this happens in homology part. Is this hinting the bicomplex for spectral sequence?
abstract-algebra homological-algebra
asked Jul 17 at 23:01
user45765
2,1942718
This is a part of statement made in Cartan Eilenberg Homological Algebra Pg 44 on Satellite.
Given $0to A^0to A^1tocdotsto A^qto 0$, one can induce connecting homomorphism by short exact sequence by applying satellite construction as satellite functor induces long exact sequence. Say $T=T^n$ is a family of additive functor with connecting homomorphism $T^n(A'')to T^n+1(A')$ for any short exact sequence $0to A'to Ato A''to 0$. Consider the following commutative diagram.
$0to A'to Ato A''to 0$
$0to B'to Bto B''to 0$
$0to C'to Cto C''to 0$
There will be arrows in vertical direction s.t above diagram is commutative and exact in columns.
Then there is an anticommutative diagram resulted. The morphisms $T^n-1(C'')to T^n(C')to T^n+1(A')$ and $T^n-1(C'')to T^n(A'')to T^n+1(A')$ differ by a minus sign.
$textbfQ:$ Where is the origin of this anticommutative diagram? Any particular reason to expect this? I am aware this happens in homology part. Is this hinting the bicomplex for spectral sequence?
abstract-algebra homological-algebra
asked Jul 17 at 23:01
user45765
2,1942718
asked Jul 17 at 23:01
user45765
2,1942718
asked Jul 17 at 23:01
user45765
2,1942718
asked Jul 17 at 23:01
user45765
2,1942718
2,1942718
add a comment |Â
add a comment |Â
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