Mixing
How do you split a long exact sequence into short exact sequences?
$begingroup$
How does one split up a long exact sequence into short exact sequences?
Say you have some longs exact sequences of modules
$$
0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots
$$
I've read it's possible to split this into short exact sequences. What exactly does that mean? Would it be written as short exact sequences, one appended to another like
$$
0longrightarrow N_1longrightarrow M_1longrightarrow N'_1longrightarrow 0longrightarrow N_2longrightarrow M_2longrightarrow N'_2longrightarrow 0 longrightarrowcdots?
$$
If so, how does this work? Merci.
abstract-algebra
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
$endgroup$
add a comment |
$begingroup$
How does one split up a long exact sequence into short exact sequences?
Say you have some longs exact sequences of modules
$$
0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots
$$
I've read it's possible to split this into short exact sequences. What exactly does that mean? Would it be written as short exact sequences, one appended to another like
$$
0longrightarrow N_1longrightarrow M_1longrightarrow N'_1longrightarrow 0longrightarrow N_2longrightarrow M_2longrightarrow N'_2longrightarrow 0 longrightarrowcdots?
$$
If so, how does this work? Merci.
abstract-algebra
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
$endgroup$
add a comment |
$begingroup$
How does one split up a long exact sequence into short exact sequences?
Say you have some longs exact sequences of modules
$$
0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots
$$
I've read it's possible to split this into short exact sequences. What exactly does that mean? Would it be written as short exact sequences, one appended to another like
$$
0longrightarrow N_1longrightarrow M_1longrightarrow N'_1longrightarrow 0longrightarrow N_2longrightarrow M_2longrightarrow N'_2longrightarrow 0 longrightarrowcdots?
$$
If so, how does this work? Merci.
abstract-algebra
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
$endgroup$
How does one split up a long exact sequence into short exact sequences?
Say you have some longs exact sequences of modules
$$
0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots
$$
I've read it's possible to split this into short exact sequences. What exactly does that mean? Would it be written as short exact sequences, one appended to another like
$$
0longrightarrow N_1longrightarrow M_1longrightarrow N'_1longrightarrow 0longrightarrow N_2longrightarrow M_2longrightarrow N'_2longrightarrow 0 longrightarrowcdots?
$$
If so, how does this work? Merci.
abstract-algebra
abstract-algebra
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
asked Jan 23 '12 at 8:29
GGGGGGGG
15324
15324
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can think of the long exact sequence
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots$$
as a collection of short exact sequences
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}mathrm{Image}(phi_2)longrightarrow 0$$
$$0longrightarrowmathrm{Coker}(phi_2)stackrel{phi_3}{longrightarrow} M_4stackrel{phi_4}{longrightarrow}mathrm{Image}(phi_4)longrightarrow 0$$
$$vdots$$
where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
$endgroup$
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
add a comment |
$begingroup$
Just to expand on @AlexBecker's answer above.
Let's say we have a long exact sequence of $A$-modules
$$dots to M_{i-1} xrightarrow{f_i} M_i xrightarrow{f_{i+1}} M_{i+1} dots$$
If we let $N_i = operatorname{Im}(f_i) = operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 to N_i xrightarrow{iota} M_i xrightarrow{pi} M_i / N_i to 0 (*)$$
where $iota : N_i to M_i$ is the inclusion map and $pi : M_i to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.
Now since $N_{i+1} = operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/operatorname{ker}(f_{i+1}) cong operatorname{Im}(f_{i+1}) iff M_i/N_i cong N_{i+1}$. If we let $f : M_i/N_i to N_{i+1}$ be this isomorphism then letting $pi' = f circ pi$ we obtain a short exact sequence
$$0 to N_i xrightarrow{iota} M_i xrightarrow{pi'} N_{i+1} to 0 (**)$$
which we can then rewrite by definition of the $N_i$'s as
$$0 to operatorname{Im}(f_i) xrightarrow{iota} M_i xrightarrow{pi'} operatorname{Im}(f_{i+1}) to 0 (**)$$
Now here's a Lemma that we will use below
Lemma: If we have $A$-modules $M, M', M''$, then $0 to M' xrightarrow{f} M xrightarrow{g} M'' to 0$ is exact $iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $operatorname{Coker}(f)$ onto $M''$
Replacing $pi'$ in the exact sequence $(**)$ by $pi''$ which is the isomorphism induced by $pi'$ of $operatorname{Coker}(iota)$ onto $N_{i+1}$ we obtain the following short exact sequence
$$0 to operatorname{ker}(f_{i+1}) xrightarrow{iota} M_i xrightarrow{pi''} operatorname{Coker}(iota) to 0 (***)$$
So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
$endgroup$
add a comment |
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2 Answers
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$begingroup$
You can think of the long exact sequence
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots$$
as a collection of short exact sequences
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}mathrm{Image}(phi_2)longrightarrow 0$$
$$0longrightarrowmathrm{Coker}(phi_2)stackrel{phi_3}{longrightarrow} M_4stackrel{phi_4}{longrightarrow}mathrm{Image}(phi_4)longrightarrow 0$$
$$vdots$$
where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
$endgroup$
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
add a comment |
$begingroup$
You can think of the long exact sequence
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots$$
as a collection of short exact sequences
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}mathrm{Image}(phi_2)longrightarrow 0$$
$$0longrightarrowmathrm{Coker}(phi_2)stackrel{phi_3}{longrightarrow} M_4stackrel{phi_4}{longrightarrow}mathrm{Image}(phi_4)longrightarrow 0$$
$$vdots$$
where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
$endgroup$
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
add a comment |
$begingroup$
You can think of the long exact sequence
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots$$
as a collection of short exact sequences
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}mathrm{Image}(phi_2)longrightarrow 0$$
$$0longrightarrowmathrm{Coker}(phi_2)stackrel{phi_3}{longrightarrow} M_4stackrel{phi_4}{longrightarrow}mathrm{Image}(phi_4)longrightarrow 0$$
$$vdots$$
where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
$endgroup$
You can think of the long exact sequence
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}M_3stackrel{phi_3}{longrightarrow}M_4stackrel{phi_4}{longrightarrow}cdots$$
as a collection of short exact sequences
$$0longrightarrow M_1stackrel{phi_1}{longrightarrow}M_2stackrel{phi_2}{longrightarrow}mathrm{Image}(phi_2)longrightarrow 0$$
$$0longrightarrowmathrm{Coker}(phi_2)stackrel{phi_3}{longrightarrow} M_4stackrel{phi_4}{longrightarrow}mathrm{Image}(phi_4)longrightarrow 0$$
$$vdots$$
where each sequence after the first begins with the relevant cokernel (well, so does the first, but this is just $M_1$) and ends with the relevant image. I have abused notation here by writing $phi_n$ for the maps from the cokernel which where originally from the corresponding module; this is not a serious issue because exactness of the original sequence ensures that the natural maps (defined by sending an equivalence class to the image of a representative) will be well-defined. One could write this as a single long chain like you proposed, but I prefer not to.
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
answered Jan 23 '12 at 8:43
Alex BeckerAlex Becker
48.8k698161
48.8k698161
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
add a comment |
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Thanks Alex, I understand now.
$endgroup$
– GGGG
Jan 23 '12 at 19:52
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
$begingroup$
Is this some sort of reverse to the Snake lemma?
$endgroup$
– gary
May 27 '18 at 20:08
add a comment |
$begingroup$
Just to expand on @AlexBecker's answer above.
Let's say we have a long exact sequence of $A$-modules
$$dots to M_{i-1} xrightarrow{f_i} M_i xrightarrow{f_{i+1}} M_{i+1} dots$$
If we let $N_i = operatorname{Im}(f_i) = operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 to N_i xrightarrow{iota} M_i xrightarrow{pi} M_i / N_i to 0 (*)$$
where $iota : N_i to M_i$ is the inclusion map and $pi : M_i to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.
Now since $N_{i+1} = operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/operatorname{ker}(f_{i+1}) cong operatorname{Im}(f_{i+1}) iff M_i/N_i cong N_{i+1}$. If we let $f : M_i/N_i to N_{i+1}$ be this isomorphism then letting $pi' = f circ pi$ we obtain a short exact sequence
$$0 to N_i xrightarrow{iota} M_i xrightarrow{pi'} N_{i+1} to 0 (**)$$
which we can then rewrite by definition of the $N_i$'s as
$$0 to operatorname{Im}(f_i) xrightarrow{iota} M_i xrightarrow{pi'} operatorname{Im}(f_{i+1}) to 0 (**)$$
Now here's a Lemma that we will use below
Lemma: If we have $A$-modules $M, M', M''$, then $0 to M' xrightarrow{f} M xrightarrow{g} M'' to 0$ is exact $iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $operatorname{Coker}(f)$ onto $M''$
Replacing $pi'$ in the exact sequence $(**)$ by $pi''$ which is the isomorphism induced by $pi'$ of $operatorname{Coker}(iota)$ onto $N_{i+1}$ we obtain the following short exact sequence
$$0 to operatorname{ker}(f_{i+1}) xrightarrow{iota} M_i xrightarrow{pi''} operatorname{Coker}(iota) to 0 (***)$$
So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
$endgroup$
add a comment |
$begingroup$
Just to expand on @AlexBecker's answer above.
Let's say we have a long exact sequence of $A$-modules
$$dots to M_{i-1} xrightarrow{f_i} M_i xrightarrow{f_{i+1}} M_{i+1} dots$$
If we let $N_i = operatorname{Im}(f_i) = operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 to N_i xrightarrow{iota} M_i xrightarrow{pi} M_i / N_i to 0 (*)$$
where $iota : N_i to M_i$ is the inclusion map and $pi : M_i to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.
Now since $N_{i+1} = operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/operatorname{ker}(f_{i+1}) cong operatorname{Im}(f_{i+1}) iff M_i/N_i cong N_{i+1}$. If we let $f : M_i/N_i to N_{i+1}$ be this isomorphism then letting $pi' = f circ pi$ we obtain a short exact sequence
$$0 to N_i xrightarrow{iota} M_i xrightarrow{pi'} N_{i+1} to 0 (**)$$
which we can then rewrite by definition of the $N_i$'s as
$$0 to operatorname{Im}(f_i) xrightarrow{iota} M_i xrightarrow{pi'} operatorname{Im}(f_{i+1}) to 0 (**)$$
Now here's a Lemma that we will use below
Lemma: If we have $A$-modules $M, M', M''$, then $0 to M' xrightarrow{f} M xrightarrow{g} M'' to 0$ is exact $iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $operatorname{Coker}(f)$ onto $M''$
Replacing $pi'$ in the exact sequence $(**)$ by $pi''$ which is the isomorphism induced by $pi'$ of $operatorname{Coker}(iota)$ onto $N_{i+1}$ we obtain the following short exact sequence
$$0 to operatorname{ker}(f_{i+1}) xrightarrow{iota} M_i xrightarrow{pi''} operatorname{Coker}(iota) to 0 (***)$$
So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
$endgroup$
add a comment |
$begingroup$
Just to expand on @AlexBecker's answer above.
Let's say we have a long exact sequence of $A$-modules
$$dots to M_{i-1} xrightarrow{f_i} M_i xrightarrow{f_{i+1}} M_{i+1} dots$$
If we let $N_i = operatorname{Im}(f_i) = operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 to N_i xrightarrow{iota} M_i xrightarrow{pi} M_i / N_i to 0 (*)$$
where $iota : N_i to M_i$ is the inclusion map and $pi : M_i to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.
Now since $N_{i+1} = operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/operatorname{ker}(f_{i+1}) cong operatorname{Im}(f_{i+1}) iff M_i/N_i cong N_{i+1}$. If we let $f : M_i/N_i to N_{i+1}$ be this isomorphism then letting $pi' = f circ pi$ we obtain a short exact sequence
$$0 to N_i xrightarrow{iota} M_i xrightarrow{pi'} N_{i+1} to 0 (**)$$
which we can then rewrite by definition of the $N_i$'s as
$$0 to operatorname{Im}(f_i) xrightarrow{iota} M_i xrightarrow{pi'} operatorname{Im}(f_{i+1}) to 0 (**)$$
Now here's a Lemma that we will use below
Lemma: If we have $A$-modules $M, M', M''$, then $0 to M' xrightarrow{f} M xrightarrow{g} M'' to 0$ is exact $iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $operatorname{Coker}(f)$ onto $M''$
Replacing $pi'$ in the exact sequence $(**)$ by $pi''$ which is the isomorphism induced by $pi'$ of $operatorname{Coker}(iota)$ onto $N_{i+1}$ we obtain the following short exact sequence
$$0 to operatorname{ker}(f_{i+1}) xrightarrow{iota} M_i xrightarrow{pi''} operatorname{Coker}(iota) to 0 (***)$$
So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
$endgroup$
Just to expand on @AlexBecker's answer above.
Let's say we have a long exact sequence of $A$-modules
$$dots to M_{i-1} xrightarrow{f_i} M_i xrightarrow{f_{i+1}} M_{i+1} dots$$
If we let $N_i = operatorname{Im}(f_i) = operatorname{ker}(f_{i+1})$ for each $i$, then we obtain (for each $i$) a short exact sequence $$0 to N_i xrightarrow{iota} M_i xrightarrow{pi} M_i / N_i to 0 (*)$$
where $iota : N_i to M_i$ is the inclusion map and $pi : M_i to M_i/N_i$ is the canonical epimorphism. It shouldn't be too hard to prove that the above sequence is exact.
Now since $N_{i+1} = operatorname{Im}(f_{i+1})$, by the first isomorphism theorem we have that $M_i/operatorname{ker}(f_{i+1}) cong operatorname{Im}(f_{i+1}) iff M_i/N_i cong N_{i+1}$. If we let $f : M_i/N_i to N_{i+1}$ be this isomorphism then letting $pi' = f circ pi$ we obtain a short exact sequence
$$0 to N_i xrightarrow{iota} M_i xrightarrow{pi'} N_{i+1} to 0 (**)$$
which we can then rewrite by definition of the $N_i$'s as
$$0 to operatorname{Im}(f_i) xrightarrow{iota} M_i xrightarrow{pi'} operatorname{Im}(f_{i+1}) to 0 (**)$$
Now here's a Lemma that we will use below
Lemma: If we have $A$-modules $M, M', M''$, then $0 to M' xrightarrow{f} M xrightarrow{g} M'' to 0$ is exact $iff$ $f$ is injective, $g$ is surjective and $g$ induces an isomorphism of $operatorname{Coker}(f)$ onto $M''$
Replacing $pi'$ in the exact sequence $(**)$ by $pi''$ which is the isomorphism induced by $pi'$ of $operatorname{Coker}(iota)$ onto $N_{i+1}$ we obtain the following short exact sequence
$$0 to operatorname{ker}(f_{i+1}) xrightarrow{iota} M_i xrightarrow{pi''} operatorname{Coker}(iota) to 0 (***)$$
So these are three ways (up to isomorphism of a term in the sequence) of splitting a long exact sequence into a short exact sequence.
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
answered Jan 7 at 12:14
PerturbativePerturbative
4,23311551
4,23311551
add a comment |
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