# The Unapologetic Mathematician

## 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 $f_1$ 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.