|
|
| Line 1: |
Line 1: |
| In [[mathematics]], and more specifically in [[homological algebra]], the '''splitting lemma''' states that in any [[abelian category]], the following statements for a [[short exact sequence]] are [[Equivalence relation|equivalent]].
| | If you compare registry products there are a amount of items to look out for. Because of the sheer number of for registry products available on the Internet at the moment it could be quite simple to be scammed. Something frequently overlooked is that certain of these cleaners might in actual fact end up damaging a PC. And the registry they say they have cleaned can simply lead to more problems with the computer than the ones we started with.<br><br>Many of the reliable companies will provide a full cash back guarantee. This signifies that we have the chance to get your money back if you find the registry cleaning has not delivered what you expected.<br><br>If you compare registry cleaners you need a swiftly acting registry cleaning. It's no advantageous spending hours plus your PC waiting for your registry cleaning to complete its task. You want your cleaner to complete its task inside minutes.<br><br>The problem with many of the persons is that they do not wish to spend income. In the damaged version 1 refuses to have to pay anything and will download it from web easily. It is simple to install too. But, the issue comes whenever it is not able to detect all possible viruses, spyware and malware inside the program. This is considering it is very obsolete inside nature plus refuses to get any normal updates from the website downloaded. Thus, the program is accessible to difficulties like hacking.<br><br>To fix the problem that is caused by registry error, you have to utilize a [http://bestregistrycleanerfix.com/regzooka regzooka]. That is the safest plus simplest way for average PC consumers. But there are thousands of registry products available out there. You should find a superior 1 that could really solve your problem. If you use a terrible one, we will anticipate more difficulties.<br><br>Although I usually utilize the latest variation of browser, sometimes different extensions plus plugins become the cause of mistakes with my browser plus the system. The same is the story with my browser which was crashing frequently perhaps due to the Flash player error.<br><br>Whenever the registry is corrupt or full of mistakes, the signs could be felt by the computer owner. The slow performance, the frequent system crashes plus the nightmare of all computer owners, the blue screen of death.<br><br>By changing the means we utilize the web we can have access more of your valuable bandwidth. This may eventually provide you a quicker surfing experience. Here is a link to 3 methods to personalize a PC speed found on the net. |
| | |
| Given a short exact sequence with maps ''q'' and ''r:''
| |
| :<math>0 \rightarrow A \overset{q}{\longrightarrow} B \overset{r}{\longrightarrow} C \rightarrow 0</math>
| |
| one writes the additional arrows ''t'' and ''u'' for maps that may not exist:
| |
| :<math>0 \rightarrow A {{q \atop \longrightarrow} \atop {\longleftarrow \atop t}} B {{r \atop \longrightarrow} \atop {\longleftarrow \atop u}} C \rightarrow 0.</math>
| |
| | |
| Then the following statements are equivalent:
| |
| ;1. left split: there exists a [[Morphism|map]] ''t'': ''B'' → ''A'' such that ''tq'' is the [[identity function|identity]] on ''A'',
| |
| ;2. right split: there exists a map ''u'': ''C'' → ''B'' such that ''ru'' is the identity on ''C'',
| |
| ;3. direct sum: ''B'' is isomorphic to the [[biproduct|direct sum]] of ''A'' and ''C'', with ''q'' corresponding to the natural injection of ''A'' and ''r'' corresponding to the natural projection onto ''C''.
| |
| | |
| | |
| | |
| The short exact sequence is called ''split'' if any of the above statements hold.
| |
| | |
| (The word "map" refers to morphisms in the abelian category we are working in, not mappings between sets.)
| |
| | |
| It allows one to refine the [[first isomorphism theorem]]:
| |
| * the first isomorphism theorem states that in the above short exact sequence, <math>C \cong B/q(A)</math>
| |
| * if the sequence splits, then <math>B = q(A) \oplus u(C) \cong A \oplus C</math>, and the first isomorphism theorem is just the projection onto ''C''.
| |
| | |
| It is a categorical generalization of the [[rank–nullity theorem]] (in the form <math>V \approx \ker T \oplus \operatorname{im}\,T</math>) in [[linear algebra]].
| |
| | |
| ==Proof==
| |
| First, to show that (3) implies both (1) and (2), we assume (3) and take as ''t'' the natural projection of the direct sum onto ''A'', and take as ''u'' the natural injection of ''C'' into the direct sum.
| |
| | |
| To prove that (1) implies (3), first note that any member of ''B'' is in the set ([[Kernel (algebra)|ker]] ''t'' + [[Image_(function)|im]] ''q''). This follows since for all ''b'' in ''B'', ''b'' = (''b'' - ''qt''(''b'')) + ''qt''(''b''); ''qt''(''b'') is obviously in im ''q'', and (''b'' - ''qt''(''b'')) is in ker ''t'', since
| |
| :''t''(''b'' - ''qt''(''b'')) = ''t''(''b'') - ''tqt''(''b'') = ''t''(''b'') - (''tq'')''t''(''b'') = ''t''(''b'') - ''t''(''b'') = 0.
| |
| | |
| Next, the intersection of im ''q'' and ker ''t'' is 0, since if there exists ''a'' in ''A'' such that ''q''(''a'') = ''b'', and ''t''(''b'') = 0, then 0 = ''tq''(''a'') = ''a''; and therefore, ''b'' = 0.
| |
| | |
| This proves that ''B'' is the direct sum of im ''q'' and ker ''t''. So, for all ''b'' in ''B'', ''b'' can be uniquely identified by some ''a'' in ''A'', ''k'' in ker ''t'', such that ''b'' = ''q''(''a'') + ''k''. | |
| | |
| By exactness ker ''r'' = im ''q''. The subsequence ''B'' → ''C'' → 0 implies that ''r'' is onto; therefore for any ''c'' in ''C'' there exists some ''b'' = ''q''(''a'') + ''k'' such that ''c'' = ''r''(''b'') = ''r''(''q''(''a'') + ''k'') = ''r''(''k''). Therefore, for any ''c'' in ''C'', exists ''k'' in ker ''t'' such that ''c'' = ''r''(''k''), and ''r''(ker ''t'') = ''C''.
| |
| | |
| If ''r''(''k'') = 0, then ''k'' is in im ''q''; since the intersection of im ''q'' and ker ''t'' = 0, then ''k'' = 0. Therefore the restriction of the morphism ''r'' : ker ''t'' → ''C'' is an isomorphism; and ker ''t'' is isomorphic to ''C''.
| |
| | |
| Finally, im ''q'' is isomorphic to ''A'' due to the exactness of 0 → ''A'' → ''B''; so ''B'' is isomorphic to the direct sum of ''A'' and ''C'', which proves (3).
| |
| | |
| To show that (2) implies (3), we follow a similar argument. Any member of ''B'' is in the set ker ''r'' + im ''u''; since for all ''b'' in ''B'', ''b'' = (''b'' - ''ur''(''b'')) + ''ur''(''b''), which is in ker ''r'' + im ''u''. The intersection of ker ''r'' and im ''u'' is 0, since if ''r''(''b'') = 0 and ''u''(''c'') = ''b'', then 0 = ''ru''(''c'') = ''c''. | |
| | |
| By exactness, im ''q'' = ker ''r'', and since ''q'' is an injection, im ''q'' is isomorphic to ''A'', so ''A'' is isomorphic to ker ''r''. Since ''ru'' is a bijection, ''u'' is an injection, and thus im ''u'' is isomorphic to ''C''. So ''B'' is again the direct sum of ''A'' and ''C''.
| |
| | |
| ==Non-abelian groups==
| |
| In the form stated here, the splitting lemma does not hold in the full [[category of groups]], which is not an abelian category.
| |
| | |
| ===Partially true===
| |
| It is partially true: if a short exact sequence of groups is left split or a direct sum (conditions 1 or 3), then all of the conditions hold. For a direct sum this is clear, as one can inject from or project to the summands. For a left split sequence, the map <math>t \times r\colon B \to A \times C</math> gives an isomorphism, so ''B'' is a direct sum (condition 3), and thus inverting the isomorphism and composing with the natural injection <math>C \to A \times C</math> gives an injection <math>C \to B</math> splitting ''r'' (condition 2).
| |
| | |
| However, if a short exact sequence of groups is right split (condition 2), then it need not be left split or a direct sum (neither condition 1 nor 3 follows): the problem is that the image of the right splitting need not be normal. What is true in this case is that ''B'' is a [[semidirect product]], though not in general a direct product.
| |
| | |
| ===Counterexample===
| |
| To form a counterexample, take the smallest non-abelian group <math>B\cong S_3</math>, the symmetric group on three letters. Let ''A'' denote the alternating subgroup, and let <math>C=B/A\cong\{\pm 1\}</math>. Let ''q'' and ''r'' denote the inclusion map and the [[Parity of a permutation|sign]] map respectively, so that
| |
| :<math>0 \rightarrow A \stackrel{q}{\longrightarrow} B \stackrel{r}{\longrightarrow} C \rightarrow 0 \,</math>
| |
| is a short exact sequence. Condition (3) fails, because <math>S_3</math> is not abelian. But condition (2) holds: we may define ''u'': ''C'' → ''B'' by mapping the generator to any two-cycle. Note for completeness that condition (1) fails: any map ''t'': ''B'' → ''A'' must map every two-cycle to the identity because the map has to be a [[group homomorphism]], while the order of a two-cycle is 2 which can not be divided by the order of the elements in A other than the identity element, which is 3 as A is the alternating subgroup of <math>S_3</math>, or namely the cyclic group of order 3. But every permutation is a product of two-cycles, so ''t'' is the trivial map, whence ''tq'': ''A'' → ''A'' is the trivial map, not the identity.
| |
| | |
| | |
| ==References==
| |
| * [[Saunders Mac Lane]]: ''Homology''. Reprint of the 1975 edition, Springer Classics in Mathematics, ISBN 3-540-58662-8, p.16
| |
| * [[Allen Hatcher]]: ''Algebraic Topology''. 2002, Cambridge University Press, ISBN 0-521-79540-0, p.147
| |
| | |
| | |
| | |
| {{DEFAULTSORT:Splitting Lemma}}
| |
| [[Category:Homological algebra]]
| |
| [[Category:Lemmas]]
| |
| [[Category:Articles containing proofs]]
| |
If you compare registry products there are a amount of items to look out for. Because of the sheer number of for registry products available on the Internet at the moment it could be quite simple to be scammed. Something frequently overlooked is that certain of these cleaners might in actual fact end up damaging a PC. And the registry they say they have cleaned can simply lead to more problems with the computer than the ones we started with.
Many of the reliable companies will provide a full cash back guarantee. This signifies that we have the chance to get your money back if you find the registry cleaning has not delivered what you expected.
If you compare registry cleaners you need a swiftly acting registry cleaning. It's no advantageous spending hours plus your PC waiting for your registry cleaning to complete its task. You want your cleaner to complete its task inside minutes.
The problem with many of the persons is that they do not wish to spend income. In the damaged version 1 refuses to have to pay anything and will download it from web easily. It is simple to install too. But, the issue comes whenever it is not able to detect all possible viruses, spyware and malware inside the program. This is considering it is very obsolete inside nature plus refuses to get any normal updates from the website downloaded. Thus, the program is accessible to difficulties like hacking.
To fix the problem that is caused by registry error, you have to utilize a regzooka. That is the safest plus simplest way for average PC consumers. But there are thousands of registry products available out there. You should find a superior 1 that could really solve your problem. If you use a terrible one, we will anticipate more difficulties.
Although I usually utilize the latest variation of browser, sometimes different extensions plus plugins become the cause of mistakes with my browser plus the system. The same is the story with my browser which was crashing frequently perhaps due to the Flash player error.
Whenever the registry is corrupt or full of mistakes, the signs could be felt by the computer owner. The slow performance, the frequent system crashes plus the nightmare of all computer owners, the blue screen of death.
By changing the means we utilize the web we can have access more of your valuable bandwidth. This may eventually provide you a quicker surfing experience. Here is a link to 3 methods to personalize a PC speed found on the net.