Editing Splitting lemma (section)

=== {{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).