show that functor $L_0T$ is right exact
$begingroup$
Show that $L_0T$ is right exact.Here T is an additive functor from category of $Lambda$-modules to abelian group.
I try to prove it,but I think the condition is too little.Any hints?
homological-algebra
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
$endgroup$
add a comment |
$begingroup$
Show that $L_0T$ is right exact.Here T is an additive functor from category of $Lambda$-modules to abelian group.
I try to prove it,but I think the condition is too little.Any hints?
homological-algebra
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
$endgroup$
add a comment |
$begingroup$
Show that $L_0T$ is right exact.Here T is an additive functor from category of $Lambda$-modules to abelian group.
I try to prove it,but I think the condition is too little.Any hints?
homological-algebra
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
$endgroup$
Show that $L_0T$ is right exact.Here T is an additive functor from category of $Lambda$-modules to abelian group.
I try to prove it,but I think the condition is too little.Any hints?
homological-algebra
homological-algebra
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
asked Jan 1 at 8:22
Daniel XuDaniel Xu
727
727
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
By construction, if you have a short exact sequence $0to X'to Xto X''to 0$, you get a long exact sequence whose initial terms look like $cdotsto L_1TX'' to L_0TX'to L_0TXto L_0TX''to 0$ so $L_0T$ is right exact.
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
$endgroup$
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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$begingroup$
By construction, if you have a short exact sequence $0to X'to Xto X''to 0$, you get a long exact sequence whose initial terms look like $cdotsto L_1TX'' to L_0TX'to L_0TXto L_0TX''to 0$ so $L_0T$ is right exact.
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
$endgroup$
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
add a comment |
$begingroup$
By construction, if you have a short exact sequence $0to X'to Xto X''to 0$, you get a long exact sequence whose initial terms look like $cdotsto L_1TX'' to L_0TX'to L_0TXto L_0TX''to 0$ so $L_0T$ is right exact.
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
$endgroup$
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
add a comment |
$begingroup$
By construction, if you have a short exact sequence $0to X'to Xto X''to 0$, you get a long exact sequence whose initial terms look like $cdotsto L_1TX'' to L_0TX'to L_0TXto L_0TX''to 0$ so $L_0T$ is right exact.
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
$endgroup$
By construction, if you have a short exact sequence $0to X'to Xto X''to 0$, you get a long exact sequence whose initial terms look like $cdotsto L_1TX'' to L_0TX'to L_0TXto L_0TX''to 0$ so $L_0T$ is right exact.
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
answered Jan 2 at 1:06
Pedro Tamaroff♦Pedro Tamaroff
97.1k10153297
97.1k10153297
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
add a comment |
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
but by definition of right exact,the given sequence is $X' to X to X'' to 0$
$endgroup$
– Daniel Xu
Jan 2 at 3:39
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
$begingroup$
@DanielXu The definitions are equivalent, you can always assume the first map is a monomorphism.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 16:21
add a comment |
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