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.
Like this:
Like Loading...
Related
October 1, 2007 -
Posted by John Armstrong |
Category theory
No comments yet.
Leave a Reply