Mixing
A kind of horseshoe lemma on the injective resolution
$begingroup$
We know that in an appropriate setting, with some exact sequence $0 to A to B to C to 0$ of $R$-modules with suitable $R$ (commutative ring with unity), we have injective resolution of $A,C$. Then, the Horseshoe lemma says that we have injective resolution of $B$ which makes every diagram commutes. Then, we can take $H^{n}$ functor to get a long exact sequence.
My question is, if we know injective resolution of $B$ and $C$, which already commutes, is it possible to get an injective resolution of $A$ which commutes with injective resolution of $B$? Furthermore, from the long exact sequence of cohomology, can we calculate $H^{n}(A)$ from the information of $H^{n-1}(C)$ and $H^{n}(B)$?
homology-cohomology homological-algebra injective-module
asked Jan 25 at 20:58
user124697user124697
638515
$endgroup$
add a comment |
$begingroup$
We know that in an appropriate setting, with some exact sequence $0 to A to B to C to 0$ of $R$-modules with suitable $R$ (commutative ring with unity), we have injective resolution of $A,C$. Then, the Horseshoe lemma says that we have injective resolution of $B$ which makes every diagram commutes. Then, we can take $H^{n}$ functor to get a long exact sequence.
My question is, if we know injective resolution of $B$ and $C$, which already commutes, is it possible to get an injective resolution of $A$ which commutes with injective resolution of $B$? Furthermore, from the long exact sequence of cohomology, can we calculate $H^{n}(A)$ from the information of $H^{n-1}(C)$ and $H^{n}(B)$?
homology-cohomology homological-algebra injective-module
asked Jan 25 at 20:58
user124697user124697
638515
$endgroup$
1
$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25
add a comment |
$begingroup$
We know that in an appropriate setting, with some exact sequence $0 to A to B to C to 0$ of $R$-modules with suitable $R$ (commutative ring with unity), we have injective resolution of $A,C$. Then, the Horseshoe lemma says that we have injective resolution of $B$ which makes every diagram commutes. Then, we can take $H^{n}$ functor to get a long exact sequence.
My question is, if we know injective resolution of $B$ and $C$, which already commutes, is it possible to get an injective resolution of $A$ which commutes with injective resolution of $B$? Furthermore, from the long exact sequence of cohomology, can we calculate $H^{n}(A)$ from the information of $H^{n-1}(C)$ and $H^{n}(B)$?
homology-cohomology homological-algebra injective-module
asked Jan 25 at 20:58
user124697user124697
638515
$endgroup$
We know that in an appropriate setting, with some exact sequence $0 to A to B to C to 0$ of $R$-modules with suitable $R$ (commutative ring with unity), we have injective resolution of $A,C$. Then, the Horseshoe lemma says that we have injective resolution of $B$ which makes every diagram commutes. Then, we can take $H^{n}$ functor to get a long exact sequence.
My question is, if we know injective resolution of $B$ and $C$, which already commutes, is it possible to get an injective resolution of $A$ which commutes with injective resolution of $B$? Furthermore, from the long exact sequence of cohomology, can we calculate $H^{n}(A)$ from the information of $H^{n-1}(C)$ and $H^{n}(B)$?
homology-cohomology homological-algebra injective-module
homology-cohomology homological-algebra injective-module
asked Jan 25 at 20:58
user124697user124697
638515
asked Jan 25 at 20:58
user124697user124697
638515
asked Jan 25 at 20:58
user124697user124697
638515
asked Jan 25 at 20:58
user124697user124697
638515
asked Jan 25 at 20:58
user124697user124697
638515
638515
1
$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25
add a comment |
1
$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25
1
1
$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25
add a comment |
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$begingroup$
what we want to get is that there is a short exact sequence of injective resolutions:$0rightarrow I(A)rightarrow I(B)rightarrow I(C)rightarrow 0$.it is clear that you can get injective resolution of A which commtes with injective resolution of B by comparison theorem.
$endgroup$
– Sky
Jan 26 at 2:02
$begingroup$
@Sky Oh yes, you right. Comparison theorem makes it possible. Thanks!
$endgroup$
– user124697
Jan 27 at 17:25