nLab
five lemma

Idea

The five lemma is one of the basic lemmas of homological algebra, useful for example in the construction of the connecting homomorphism in the homology long exact sequence.

Five lemma

Consider a commutative diagram in a fixed abelian category of the form

$$\begin{array}{ccccccccc}{A}_1& \to & A_2& \to & A_3& \to & A_4& \to & A_5\\ \downarrow f_1& & \downarrow f_2& & \downarrow f_3& & \downarrow f_4& & \downarrow f_5\\ B_1& \to & B_2& \to & B_3& \to & B_4& \to & B_5\end{array}$$\array{ A_1 & \to & A_2 & \to & A_3 & \to & A_4 &\to & A_5\\ \downarrow f_1 &&\downarrow f_2 &&\downarrow f_3 &&\downarrow f_4 &&\downarrow f_5 \\ B_1 & \to & B_2 & \to & B_3 & \to & B_4 &\to & B_5 }

where the top and bottom are exact sequences. For simplicity we denote all the differentials in both exact sequences by $d$.

Lemma

(five lemma)

• 1) If $f_2$ and $f_4$ are epis and $f_5$ is mono, then $f_3$ is epi.

• 2) If $f_2$ and $f_4$ are monic and $f_1$ is epi, then $f_3$ is mono.

• 1+2) If $f_2$ and $f_4$ are isos, $f_1$ is epi, and $f_5$ is mono, then $f_3$ is iso.

Proof

We can embed into a category of left modules over a fixed ring (though this requires the category to be small, one can always take a smaller abelian subcategory containing the morphism in the diagram which is small). Then we can do the diagram chasing using elements in that setup. We prove only 1) as 2) is dual.

The proof of 1) is by contradiction. Suppose $f_3$ is not epi, hence there is $b\in B_3$ which is not in the image of $f_3$. Since $f_4$ is epi, one can choose an element $a_4\in A_4$ such that $f_4(a_4)=d(b)$. Now $$0=d^{2}b=d{f}_4(a_4)=f_{5}d(a_4)$. Since $f_5$ is monic that means that $d{a}_4=0$ as well. By the exactness of the upper row, that means there is $a_3\in A_3$ such that $d{a}_3=a_4$, hence also $d{f}_3(a_3)=f_{4}d(a_3)=f_4(a_4)=db$. We would like that $f_3(a_3)$ be equal to $b$ but this is not so, we just see that $d(b-f_3(a_3))=0$ and hence by exactness of the lower row there is $b\prime \in B_2$ such that $db\prime =b-f_3(a_3)$. Since $f_2$ is also epi, there is $a_2\in A_2$ such that $f_2(a_2)=b\prime$. Now $d{a}_2+a_3\in A_3$ is such that

$$f_3(d{a}_2+a_3)={\mathrm{df}}_2(a_2)+f_3(a_3)=\mathrm{db}\prime +f_3(a_3)=b-f_3(a_3)+f_3(a_3)=b$$f_3 (d a_2 + a_3) = df_2(a_2)+f_3(a_3) = db'+f_3(a_3) = b - f_3(a_3)+f_3(a_3) = b

with contradiction.

Short split five lemma

The special case of five lemma is a short five lemma where $A_1,B_1,A_5,B_5$ are all zero objects. It may hold in more general setups, sometimes with additional assumptions.

The short split five lemma is a statement usually stated in the setup of semiabelian categories:

Given a commutative diagram

$$\begin{array}{ccccc}L& \stackrel{l}{\to }& H& \stackrel{q}{\to }& C\\ {}^u\downarrow & & {\downarrow }^w& & {\downarrow }^v\\ K& \underset{k}{\to }& G& \underset{p}{\to }& B\end{array}$$\array{L & \overset{l}{\to} & H & \overset{q}{\to} & C\\ ^u\downarrow && \downarrow^w && \downarrow^v \\ K & \underset{k}{\to} & G& \underset{p}{\to} & B}

where $p$ and $q$ are split epimorphisms and $l$ and $k$ are their kernels, if $u$ and $v$ are isomorphisms then so is $w$.