The Five Lemma
Today I’ll prove the “five lemma”, using the diagram chasing rules I outlined last time. This is an extension of the short five lemma from last week.
We consider the following commutative diagram with exact rows:
The five lemma states that if is epic, and
and
are monic, then
is monic. By our chasing rules this means that for all
,
implies that
. So let’s start with a member
of
and assume that
.
First off we can tell that , and since
is monic this means
. Now the upper row is exact at
, so there must exist a member
with
. We find
, so by the exactness of the lower row at
there must be a member
with
. Because
is epic, there is a member
with
. Then
, and so
. Then
, but exactness tells us that
, which leads us to conclude that
. Therefore
is monic.
Dually, if and
are epic while
is monic we conclude that
is epic. And then if
is epic,
and
are isomorphisms, and
is monic, then
is an isomorphism.
It’s instructive to run through the above chase in both languages. First read as denoting an element of an abelian group, while
denotes equality of elements. Then go back and reconstruct the proof in terms of morphisms between objects of general abelian categories. Thinking of members as behaving like actual elements allows us to create a proof that works for any abelian category we happen to be considering.
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