Proof of the Five Lemma

We give a complete proof of the Five Lemma by diagram chasing in $R\text{-}\mathbf{Mod}$ .

Setup

Consider the commutative diagram with exact rows:

\begin{array}{ccccccccc} A_1 & \xrightarrow{d_1} & A_2 & \xrightarrow{d_2} & A_3 & \xrightarrow{d_3} & A_4 & \xrightarrow{d_4} & A_5 \\[6pt] \downarrow\scriptstyle{\alpha_1} & & \downarrow\scriptstyle{\alpha_2} & & \downarrow\scriptstyle{\alpha_3} & & \downarrow\scriptstyle{\alpha_4} & & \downarrow\scriptstyle{\alpha_5} \\[6pt] B_1 & \xrightarrow{e_1} & B_2 & \xrightarrow{e_2} & B_3 & \xrightarrow{e_3} & B_4 & \xrightarrow{e_4} & B_5 \end{array}

We prove the two "Four Lemma" halves separately.

Part 1: $\alpha_3$ is Monic

Proofalpha_3 is monic (assuming alpha_2, alpha_4 monic and alpha_1 epic)

Assume $\alpha_2$ and $\alpha_4$ are monic and $\alpha_1$ is epic. We show $\alpha_3$ is monic.

Let $a_3 \in A_3$ with $\alpha_3(a_3) = 0$ . We need to show $a_3 = 0$ .

Step 1. Compute $d_3(a_3) \in A_4$ . We have $\alpha_4(d_3(a_3)) = e_3(\alpha_3(a_3)) = e_3(0) = 0$ . Since $\alpha_4$ is monic, $d_3(a_3) = 0$ .

Step 2. By exactness of the top row at $A_3$ : $\ker d_3 = \mathrm{im}\, d_2$ . So $a_3 = d_2(a_2)$ for some $a_2 \in A_2$ .

Step 3. Compute $e_2(\alpha_2(a_2)) = \alpha_3(d_2(a_2)) = \alpha_3(a_3) = 0$ . By exactness of the bottom row at $B_2$ : $\alpha_2(a_2) \in \ker e_2 = \mathrm{im}\, e_1$ . So $\alpha_2(a_2) = e_1(b_1)$ for some $b_1 \in B_1$ .

Step 4. Since $\alpha_1$ is epic, $b_1 = \alpha_1(a_1)$ for some $a_1 \in A_1$ . Then $\alpha_2(a_2) = e_1(\alpha_1(a_1)) = \alpha_2(d_1(a_1))$ .

Step 5. Since $\alpha_2$ is monic, $a_2 = d_1(a_1)$ . Therefore $a_3 = d_2(a_2) = d_2(d_1(a_1))$ .

Step 6. By exactness of the top row at $A_2$ : $\mathrm{im}\, d_1 \subseteq \ker d_2$ . So $d_2(d_1(a_1)) = 0$ . Thus $a_3 = 0$ .

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Part 2: $\alpha_3$ is Epic

Proofalpha_3 is epic (assuming alpha_2, alpha_4 epic and alpha_5 monic)

Assume $\alpha_2$ and $\alpha_4$ are epic and $\alpha_5$ is monic. We show $\alpha_3$ is epic.

Let $b_3 \in B_3$ . We need to find $a_3 \in A_3$ with $\alpha_3(a_3) = b_3$ .

Step 1. Compute $e_3(b_3) \in B_4$ . Since $\alpha_4$ is epic, $e_3(b_3) = \alpha_4(a_4)$ for some $a_4 \in A_4$ .

Step 2. Compute $\alpha_5(d_4(a_4)) = e_4(\alpha_4(a_4)) = e_4(e_3(b_3)) = 0$ (by exactness of the bottom row: $e_4 \circ e_3 = 0$ ). Since $\alpha_5$ is monic, $d_4(a_4) = 0$ .

Step 3. By exactness of the top row at $A_4$ : $a_4 \in \ker d_4 = \mathrm{im}\, d_3$ . So $a_4 = d_3(a_3')$ for some $a_3' \in A_3$ .

Step 4. Compute $e_3(b_3 - \alpha_3(a_3')) = e_3(b_3) - e_3(\alpha_3(a_3')) = \alpha_4(a_4) - \alpha_4(d_3(a_3')) = \alpha_4(a_4 - d_3(a_3')) = \alpha_4(0) = 0$ .

Step 5. So $b_3 - \alpha_3(a_3') \in \ker e_3 = \mathrm{im}\, e_2$ . Write $b_3 - \alpha_3(a_3') = e_2(b_2)$ for some $b_2 \in B_2$ .

Step 6. Since $\alpha_2$ is epic, $b_2 = \alpha_2(a_2)$ for some $a_2 \in A_2$ . Then $b_3 - \alpha_3(a_3') = e_2(\alpha_2(a_2)) = \alpha_3(d_2(a_2))$ .

Step 7. Set $a_3 = a_3' + d_2(a_2)$ . Then $\alpha_3(a_3) = \alpha_3(a_3') + \alpha_3(d_2(a_2)) = \alpha_3(a_3') + (b_3 - \alpha_3(a_3')) = b_3$ .

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Conclusion

ProofFull Five Lemma

If $\alpha_1, \alpha_2, \alpha_4, \alpha_5$ are all isomorphisms, then:

$\alpha_1$ is epic, $\alpha_2$ and $\alpha_4$ are monic $\Rightarrow$ $\alpha_3$ is monic (Part 1).
$\alpha_5$ is monic, $\alpha_2$ and $\alpha_4$ are epic $\Rightarrow$ $\alpha_3$ is epic (Part 2).

Therefore $\alpha_3$ is an isomorphism. $\square$

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Remarks

RemarkElement-free proof

A proof using only the universal properties of kernels and cokernels (without elements) is possible but more involved. The element-chasing proof is valid in any abelian category by the Freyd-Mitchell Embedding Theorem.

ExampleApplication: comparing long exact sequences

The Five Lemma is most commonly applied when comparing two long exact sequences connected by a morphism. If the maps on four out of five terms are known to be isomorphisms (e.g., by induction or by a separate argument), the Five Lemma gives the fifth for free.

ExampleApplication: quasi-isomorphism from mapping cone

A chain map $f$ is a quasi-isomorphism iff $\mathrm{Cone}(f)$ is acyclic. This follows from the long exact sequence of the mapping cone and the Five Lemma: if $H^n(\mathrm{Cone}(f)) = 0$ for all $n$ , then $H^n(f)$ is an isomorphism for all $n$ .

ExampleShort Five Lemma

The Short Five Lemma is the special case where $A_1 = B_1 = 0$ and $A_5 = B_5 = 0$ . In this case, the hypotheses simplify: if $\alpha_2$ and $\alpha_4$ are isomorphisms, then $\alpha_3$ is an isomorphism. This is the most frequently used version.

ExampleNaturality of connecting homomorphisms

The Five Lemma, combined with the Snake Lemma, shows that connecting homomorphisms in long exact sequences are natural: a morphism of short exact sequences induces a morphism of long exact sequences, and the connecting homomorphism commutes with the induced maps.