diagram chasing in homological algebra
salamander lemma $\Rightarrow$
four lemma $\Rightarrow$ five lemma
snake lemma $\Rightarrow$ connecting homomorphism
(also nonabelian homological algebra)
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.
Let $\mathcal{A}$ be an abelian category. Consider a commutative diagram in $\mathcal{A}$ of the form
where the top and bottom rows are exact sequences. For simplicity we denote all the differentials in both exact sequences by $d$.
sharp five lemma (essentially the weak four lemma)
(weak) five lemma (conjunction of the two statements above)
If $f_2$ and $f_4$ are isos, $f_1$ is epi, and $f_5$ is mono, then $f_3$ is iso.
on terminology
The weak four lemma is another terminology (cf. MacLane, Homology) for the same as 1.1 and 1.2 except that in 1.1 $f_1$ is not required to exist, and in 1.2 $f_5$ is not required to exist (see four lemma), where the dropped requirements are inessential as not used in the proof.
The four lemma follows immediately from the salamander lemma, as discussed at salamander lemma - impliciations - four lemma. Here is direct proof.
By the Freyd-Mitchell embedding theorem we can always assume that the abelian category is $R$Mod (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.
Suppose $b\in B_3$. 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 a monomorphism 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 $d f_3 (a_3) = f_4 d (a_3) = f_4(a_4) = d b$. 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 $b'\in B_2$ such that $d b' = b-f_3(a_3)$. Since $f_2$ is also epi, there is $a_2\in A_2$ such that $f_2(a_2) = b'$. Now $d a_2+a_3\in A_3$ is such that
demonstrating that $b$ is in the image of $f_3$.
Hence $f_3$ is an epimorphism.
The five lemma also holds in the category Grp of groups, by essentially the same diagram-chasing proof.
One can avoid appealing to the Freyd-Mitchell embedding theorem if one works with generalized elements or uses the device of interpreting regular logic in the given abelian category. The former requires a bit of manual reformulation, while the latter is almost automatic, as the element-based proof given above only uses (constructive) regular reasoning.
(short five lemma)
Let $A \to B \to C$ and $A \to \tilde B \to C$ be two exact sequences. If a homomorphism $f \colon B \to \tilde B$ makes the diagram
commute, then $f$ is an isomorphism.
Apply prop. 1 to the diagram
A special case of the five lemma is the short five lemma where the objects $A_1,B_1,A_5,B_5$ above 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:
(short split five lemma)
Given a commutative diagram
where $p$ and $q$ are split epimorphisms and $l$ and $k$ are their kernels, if $u$ and $v$ are isomorphisms then so is $w$.
The short five lemma holds in the category of abelian topological groups, even though that category is not semi-abelian. For a proof, see this paper by Borceux and Clementino.
Early references of the 5-lemma
Modern
In nonabelian context
The short 5-lemma also appears in various topological algebra contexts; see for example