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: ''See also 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 for a short exact sequence are equivalent. Given a short exact sequence with maps ''q'' and ''r:'' : one writes the additional arrows ''t'' and ''u'' for maps that may not exist: : Then the following statements are equivalent: ;1. left split: there exists a map ''t'': ''B'' → ''A'' such that ''tq'' is the 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 direct sum of ''A'' and ''C'', with ''q'' corresponding to the natural injection of ''A'' and ''r'' corresponding to the natural projection onto ''C''. More precisely, there is an isomorphism of short exact sequences between the given sequence and the sequence with ''B'' replaced by the direct sum of ''A'' and ''C'', where the maps are the canonical inclusion and projection. Just an isomorphism of ''B'' with the direct sum is not sufficient. 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, (i.e. "C" isomorphic to the coimage of "r" or cokernel of "q") * if the sequence splits, then , and the first isomorphism theorem is just the projection onto ''C''. It is a categorical generalization of the rank–nullity theorem (in the form ) 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 (ker ''t'' + 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''. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「splitting lemma」の詳細全文を読む スポンサード リンク
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