Does every short exact sequence split?Additive functor over a short split exact sequence.Splitting short exact sequence of space groupsShort exact sequence is split iff contractibleSplitting of Short Exact sequenceDetermine if the following short exact sequence is split.A short exact sequence with $M=M_1 oplus M_2$ that does not splitShort exact sequence terminating at $mathbbZ/2$Short exact sequence and splitting lemmaShort exact sequence on modulesDoes an embedding necessarily make a short exact sequence split?
Is it logically or scientifically possible to artificially send energy to the body?
Madden-Julian Oscillation (MJO) - How to interpret the index?
Am I breaking OOP practice with this architecture?
Which is the best way to check return result?
Arrow those variables!
Bullying boss launched a smear campaign and made me unemployable
How badly should I try to prevent a user from XSSing themselves?
What exploit Are these user agents trying to use?
GFCI outlets - can they be repaired? Are they really needed at the end of a circuit?
What method can I use to design a dungeon difficult enough that the PCs can't make it through without killing them?
Short story with a alien planet, government officials must wear exploding medallions
Running Low on Limestone
What does the expression "A Mann!" means
What is the most common color to indicate the input-field is disabled?
Why can't we play rap on piano?
Assassin's bullet with mercury
iPad being using in wall mount battery swollen
Personal Teleportation: From Rags to Riches
In 'Revenger,' what does 'cove' come from?
How to add frame around section using titlesec?
ssTTsSTtRrriinInnnnNNNIiinngg
What is the idiomatic way to say "clothing fits"?
If human space travel is limited by the G force vulnerability, is there a way to counter G forces?
What mechanic is there to disable a threat instead of killing it?
Does every short exact sequence split?
Additive functor over a short split exact sequence.Splitting short exact sequence of space groupsShort exact sequence is split iff contractibleSplitting of Short Exact sequenceDetermine if the following short exact sequence is split.A short exact sequence with $M=M_1 oplus M_2$ that does not splitShort exact sequence terminating at $mathbbZ/2$Short exact sequence and splitting lemmaShort exact sequence on modulesDoes an embedding necessarily make a short exact sequence split?
$begingroup$
If we have a short exact sequence $0 to A to B to C to 0$, then the map $A to B$ is injective and the map $B to C$ is surjective.
Therefore, there always exists a left inverse for $i$ and a right inverse for $j$. So, every short exact sequence splits?
abstract-algebra proof-verification algebraic-topology homological-algebra exact-sequence
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
$endgroup$
|
show 4 more comments
$begingroup$
If we have a short exact sequence $0 to A to B to C to 0$, then the map $A to B$ is injective and the map $B to C$ is surjective.
Therefore, there always exists a left inverse for $i$ and a right inverse for $j$. So, every short exact sequence splits?
abstract-algebra proof-verification algebraic-topology homological-algebra exact-sequence
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
$endgroup$
6
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
2
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
2
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
1
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
1
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36
|
show 4 more comments
$begingroup$
If we have a short exact sequence $0 to A to B to C to 0$, then the map $A to B$ is injective and the map $B to C$ is surjective.
Therefore, there always exists a left inverse for $i$ and a right inverse for $j$. So, every short exact sequence splits?
abstract-algebra proof-verification algebraic-topology homological-algebra exact-sequence
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
$endgroup$
If we have a short exact sequence $0 to A to B to C to 0$, then the map $A to B$ is injective and the map $B to C$ is surjective.
Therefore, there always exists a left inverse for $i$ and a right inverse for $j$. So, every short exact sequence splits?
abstract-algebra proof-verification algebraic-topology homological-algebra exact-sequence
abstract-algebra proof-verification algebraic-topology homological-algebra exact-sequence
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
asked Mar 28 at 19:42
Al JebrAl Jebr
4,39743478
4,39743478
6
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
2
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
2
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
1
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
1
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36
|
show 4 more comments
6
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
2
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
2
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
1
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
1
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36
6
6
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
2
2
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
2
2
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
1
1
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
1
1
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The question's already been answered in the comments, but for posterity there should be an official one.
The answer is No and a nice counterexample is
$$ 0 to mathbbZ to mathbbZ to mathbbZ/2 to 0$$
We know this can't split because the Splitting Lemma would then imply $mathbbZ cong mathbbZ oplus mathbbZ/2$, which is a contradiction because $mathbbZ$ has no torsion.
It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $mathbbZ/2 to mathbbZ$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $mathbbZ$ has no torsion.
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166321%2fdoes-every-short-exact-sequence-split%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The question's already been answered in the comments, but for posterity there should be an official one.
The answer is No and a nice counterexample is
$$ 0 to mathbbZ to mathbbZ to mathbbZ/2 to 0$$
We know this can't split because the Splitting Lemma would then imply $mathbbZ cong mathbbZ oplus mathbbZ/2$, which is a contradiction because $mathbbZ$ has no torsion.
It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $mathbbZ/2 to mathbbZ$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $mathbbZ$ has no torsion.
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
$endgroup$
add a comment |
$begingroup$
The question's already been answered in the comments, but for posterity there should be an official one.
The answer is No and a nice counterexample is
$$ 0 to mathbbZ to mathbbZ to mathbbZ/2 to 0$$
We know this can't split because the Splitting Lemma would then imply $mathbbZ cong mathbbZ oplus mathbbZ/2$, which is a contradiction because $mathbbZ$ has no torsion.
It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $mathbbZ/2 to mathbbZ$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $mathbbZ$ has no torsion.
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
$endgroup$
add a comment |
$begingroup$
The question's already been answered in the comments, but for posterity there should be an official one.
The answer is No and a nice counterexample is
$$ 0 to mathbbZ to mathbbZ to mathbbZ/2 to 0$$
We know this can't split because the Splitting Lemma would then imply $mathbbZ cong mathbbZ oplus mathbbZ/2$, which is a contradiction because $mathbbZ$ has no torsion.
It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $mathbbZ/2 to mathbbZ$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $mathbbZ$ has no torsion.
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
$endgroup$
The question's already been answered in the comments, but for posterity there should be an official one.
The answer is No and a nice counterexample is
$$ 0 to mathbbZ to mathbbZ to mathbbZ/2 to 0$$
We know this can't split because the Splitting Lemma would then imply $mathbbZ cong mathbbZ oplus mathbbZ/2$, which is a contradiction because $mathbbZ$ has no torsion.
It is true that in the category of Sets every surjective function has a right inverse and every injective function has a left inverse, but this is not true in the category of Abelian Groups. Indeed a right inverse $mathbbZ/2 to mathbbZ$ of the quotient map would be any function sending $0$ to an even number and $1$ to an odd number, but this can never be a homomorphism, again because $mathbbZ$ has no torsion.
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
edited Mar 28 at 21:54
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
answered Mar 28 at 21:37
WilliamWilliam
3,1021227
3,1021227
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3166321%2fdoes-every-short-exact-sequence-split%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
6
$begingroup$
As functions, onto maps always have inverses, but those inverses might not work as algebraic homomorphisms.
$endgroup$
– Thomas Andrews
Mar 28 at 19:46
2
$begingroup$
@ThomasAndrews That's not even a hint. That's a full answer. Please don't put those in comments.
$endgroup$
– Arthur
Mar 28 at 19:47
2
$begingroup$
As the excerpt notes, if $0to Ato Bto Cto 0$ is exact and splits, then you get $Bcong Aoplus C$. So the question you should ask yourself is, "If $B$ is an abelian group and $A$ is a subgroup, can I always write $B$ as a direct sum with $A$ as one summand?
$endgroup$
– Arturo Magidin
Mar 28 at 19:48
1
$begingroup$
You should look up the terms "projective module" and "injective module". If you have not seen modules, a quick explanation is that general R-modules have the same notion of exact sequences as abelian groups. Then, $C$ is projective, if and only if, any exact sequence $ 0 rightarrow A rightarrow B rightarrow C rightarrow 0$ splits. You might be able to guess the dual equivalence for injective modules. A great exercise would be to use this as a definition for projective abelian group and to try to characterize them.
$endgroup$
– Connor Malin
Mar 28 at 20:46
1
$begingroup$
Two things: you are correct that modules are abelian groups with extra structure, but that doesn't mean this result is stronger because the result for modules gives decompositions into direct sums of modules. The result for abelian groups does not imply it for modules (imply being used informally). In fact, the opposite is true. The result for modules implies it for abelian groups because abelian groups are $mathbbZ$ modules.
$endgroup$
– Connor Malin
Mar 29 at 3:36