Editing Splitting lemma

{{Short description|About direct sums and exact sequences}}
{{Distinguish|text=the [[splitting lemma (functions)|splitting lemma]] in [[singularity theory]]}}

In [[mathematics]], and more specifically in [[homological algebra]], the '''splitting lemma''' states that in any [[abelian category]], the following statements are [[logical equivalence|equivalent]] for a [[short exact sequence]] 
: <math>0 \longrightarrow A \mathrel{\overset{q}{\longrightarrow}} B \mathrel{\overset{r}{\longrightarrow}} C \longrightarrow 0.</math>
{{ordered list|{{glossary}}{{term|Left split}}{{defn|There exists a [[morphism]] {{math|''t'': ''B'' → ''A''}} such that {{math|''tq''}} is the [[identity function|identity]] {{math|id{{sub|''A''}}}} on {{math|''A''}},}}{{glossary end}}|{{glossary}}{{term|Right split}}{{defn|There exists a morphism {{math|''u'': ''C'' → ''B''}} such that {{math|''ru''}} is the identity {{math|id{{sub|''C''}}}} on {{math|''C''}},}}{{glossary end}}|{{glossary}}{{term|Direct sum}}{{defn|There is an [[isomorphism (category theory)|isomorphism]] {{mvar|h}} from {{math|''B''}} to the [[biproduct|direct sum]] of {{math|''A''}} and {{math|''C''}}, such that {{math|''hq''}} is the natural injection of {{math|''A''}} into the direct sum, and <math>rh^{-1}</math> is the natural projection of the direct sum [[surjective|onto]] {{math|''C''}}.}}{{glossary end}}
}}

If any of these statements holds, the sequence is called a '''[[split exact sequence]]''', and the sequence is said to ''split''.

In the above short exact sequence, where the sequence splits, it allows one to refine the [[first isomorphism theorem]], which states that:
: {{math|''C'' ≅ ''B''/ker ''r'' ≅ ''B''/''q''(''A'')}} (i.e., {{math|''C''}} isomorphic to the [[coimage]] of {{math|''r''}} or [[cokernel]] of {{math|''q''}})
to:
: {{math|''B'' {{=}} ''q''(''A'') ⊕ ''u''(''C'') ≅ ''A'' ⊕ ''C''}}
where the first isomorphism theorem is then just the projection onto {{math|''C''}}.

It is a [[category theory|categorical]] generalization of the [[rank–nullity theorem]] (in the form {{math|V ≅ ker&thinsp;''T'' ⊕ im&thinsp;''T'')}} in [[linear algebra]].

==Proof for the category of abelian groups==

=== {{math|3. ⇒ 1.}} and {{math|3. ⇒ 2.}} ===
First, to show that 3. implies both 1. and 2., we assume 3. and take as {{math|''t''}} the natural projection of the direct sum onto {{math|''A''}}, and take as {{math|''u''}} the natural injection of {{math|''C''}} into the direct sum.

=== {{math|1. ⇒ 3.}} ===
To [[mathematical proof|prove]] that 1. implies 3., first note that any member of ''B'' is in the set ({{math|[[kernel (algebra)|ker]] ''t'' + [[image (function)|im]] ''q''}}). This follows since for all {{math|''b''}} in {{math|''B''}}, {{math|''b'' {{=}} (''b'' − ''qt''(''b'')) + ''qt''(''b'')}}; {{math|''qt''(''b'')}} is in {{math|im ''q''}}, and {{math|''b'' − ''qt''(''b'')}} is in {{math|ker ''t''}}, since 
:{{math|''t''(''b'' − ''qt''(''b'')) {{=}} ''t''(''b'') − ''tqt''(''b'') {{=}} ''t''(''b'') − (''tq'')''t''(''b'') {{=}} ''t''(''b'') − ''t''(''b'') {{=}} 0.}}

Next, the [[intersection (set theory)|intersection]] of {{math|im ''q''}} and {{math|ker ''t''}} is 0, since if there exists {{math|''a''}} in {{math|''A''}} such that {{math|''q''(''a'') {{=}} ''b''}}, and {{math|''t''(''b'') {{=}} 0}}, then {{math|0 {{=}} ''tq''(''a'') {{=}} ''a''}}; and therefore, {{math|''b'' {{=}} 0}}.

This proves that {{math|''B''}} is the direct sum of {{math|im ''q''}} and {{math|ker ''t''}}. So, for all {{math|''b''}} in {{math|''B''}}, {{math|''b''}} can be uniquely identified by some {{math|''a''}} in {{math|''A''}}, {{math|''k''}} in {{math|ker ''t''}}, such that {{math|''b'' {{=}} ''q''(''a'') + ''k''}}.

By exactness {{math|ker ''r'' {{=}} im ''q''}}. The subsequence {{math|''B'' ⟶ ''C'' ⟶ 0}} implies that {{math|''r''}} is [[surjective|onto]]; therefore for any {{math|''c''}} in {{math|''C''}} there exists some {{math|''b'' {{=}} ''q''(''a'') + ''k''}} such that {{math|''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 {{math|''r''(''k'') {{=}} 0}}, then {{math|''k''}} is in {{math|im ''q''}}; since the intersection of {{math|im ''q''}} and {{math|ker ''t'' {{=}} 0}}, then {{math|''k'' {{=}} 0}}. Therefore, the [[restriction (mathematics)|restriction]] {{math|''r'': ker ''t'' → ''C''}} is an isomorphism; and {{math|ker ''t''}} is isomorphic to {{math|''C''}}.

Finally, {{math|im ''q''}} is isomorphic to {{math|''A''}} due to the exactness of {{math|0 ⟶ ''A'' ⟶ ''B''}}; so ''B'' is isomorphic to the direct sum of {{math|''A''}} and {{math|''C''}}, which proves (3).

=== {{math|2. ⇒ 3.}} ===
To show that 2. implies 3., we follow a similar argument. Any member of {{math|''B''}} is in the set {{math|ker ''r'' + im ''u''}}; since for all {{math|''b''}} in {{math|''B''}}, {{math|''b'' {{=}} (''b'' − ''ur''(''b'')) + ''ur''(''b'')}}, which is in {{math|ker ''r'' + im ''u''}}. The intersection of {{math|ker ''r''}} and {{math|im ''u''}} is {{math|0}}, since if {{math|''r''(''b'') {{=}} 0}} and {{math|''u''(''c'') {{=}} ''b''}}, then {{math|0 {{=}} ''ru''(''c'') {{=}} ''c''}}.

By exactness, {{math|im ''q'' {{=}} ker ''r''}}, and since {{math|''q''}} is an [[injective|injection]], {{math|im ''q''}} is isomorphic to {{math|''A''}}, so {{math|''A''}} is isomorphic to {{math|ker ''r''}}. Since {{math|''ru''}} is a [[bijection]], {{math|''u''}} is an injection, and thus {{math|im ''u''}} is isomorphic to {{math|''C''}}. So {{math|''B''}} is again the direct sum of {{math|''A''}} and {{math|''C''}}.

An alternative "[[abstract nonsense]]" [https://math.stackexchange.com/q/753182 proof of the splitting lemma] may be formulated entirely in [[category theory|category theoretic]] terms.

==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 (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'' × ''r'': ''B'' → ''A'' × ''C''}} gives an isomorphism, so {{math|''B''}} is a direct sum (3.), and thus inverting the isomorphism and composing with the natural injection {{math|''C'' → ''A'' × ''C''}} gives an injection {{math|''C'' → ''B''}} splitting {{math|''r''}} (2.).

However, if a short exact sequence of groups is right split (2.), then it need not be left split or a direct sum (neither 1. nor 3. follows): the problem is that the image of the right splitting need not be [[normal subgroup|normal]]. What is true in this case is that {{math|''B''}} is a [[semidirect product]], though not in general a [[direct product of groups|direct product]].

===Counterexample===
To form a counterexample, take the smallest [[non-abelian group]] {{math|''B'' ≅ ''S''{{sub|3}}}}, the [[symmetric group]] on three letters.  Let {{math|''A''}} denote the [[alternating group|alternating subgroup]], and let {{math|''C'' {{=}} ''B''/''A'' ≅ {±1}}}.  Let {{math|''q''}} and {{math|''r''}} denote the inclusion map and the [[parity of a permutation|sign]] map respectively, so that

: <math>0 \longrightarrow A \mathrel{\stackrel{q}{\longrightarrow}} B \mathrel{\stackrel{r}{\longrightarrow}} C \longrightarrow 0 </math>

is a short exact sequence.  3. fails, because {{math|''S''{{sub|3}}}} is not abelian, but 2. holds: we may define {{math|''u'': ''C'' → ''B''}} by mapping the generator to any [[cyclic permutation|two-cycle]].  Note for completeness that 1. fails: any map {{math|''t'': ''B'' → ''A''}} must map every two-cycle to the [[identity permutation|identity]] because the map has to be a [[group homomorphism]], while the [[order (group theory)|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 {{math|''A''}} is the alternating subgroup of {{math|''S''{{sub|3}}}}, or namely the [[cyclic group]] of [[order of a group|order]] 3. But every [[permutation]] is a product of two-cycles, so {{math|''t''}} is the trivial map, whence {{math|''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.&nbsp;16
* [[Allen Hatcher]]: ''Algebraic Topology''. 2002, Cambridge University Press, {{ISBN|0-521-79540-0}}, p.&nbsp;147

{{DEFAULTSORT:Splitting Lemma}}
[[Category:Homological algebra]]
[[Category:Lemmas in category theory]]
[[Category:Articles containing proofs]]