The Unapologetic Mathematician

Mathematics for the interested outsider

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:
Diagram for the Five Lemma
The five lemma states that if f_1 is epic, and f_2 and f_4 are monic, then f_3 is monic. By our chasing rules this means that for all x\in_m A_3, f_3\circ x\equiv0 implies that x\equiv0. So let’s start with a member x of A_3 and assume that f_3\circ x\equiv0.

First off we can tell that f_4\circ g_3\circ x=h_3\circ f_3\circ x\equiv0, and since f_4 is monic this means g_3\circ x\equiv0. Now the upper row is exact at A_3, so there must exist a member y\in_mA_2 with g_2\circ y\equiv x. We find h_2\circ f_2\circ y=f_3\circ g_2\circ y\equiv f_3\circ x\equiv0, so by the exactness of the lower row at B_2 there must be a member z\in_mB_1 with h_1\circ z\equiv f_2\circ y. Because f_1 is epic, there is a member w\in_mA_1 with f_1\circ w\equiv z. Then f_2\circ g_1\circ w=h_1\circ f_1\circ w\equiv h_1\circ z\equiv f_2\circ y, and so g_1\circ w\equiv y. Then x\equiv g_2\circ y\equiv g_2\circ g_1\circ w, but exactness tells us that g_2\circ g_1\equiv0, which leads us to conclude that x\equiv0. Therefore f_3 is monic.

Dually, if f_2 and f_4 are epic while f_5 is monic we conclude that f_3 is epic. And then if f_1 is epic, f_2 and f_4 are isomorphisms, and f_5 is monic, then f_3 is an isomorphism.

It’s instructive to run through the above chase in both languages. First read \in_m as denoting an element of an abelian group, while \equiv 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.

October 1, 2007 Posted by | Category theory | Leave a comment