From the 9 Lemma to the Five Lemma

From the $3\times 3$$3\times3$ Lemma to the Five Lemma

This note explains three basic results in homological algebra:

1. two-out-of-three for exact complexes in a short exact sequence of complexes;

2. the $3\times 3$$3\times3$ lemma;

3. the reduction of the five lemma to the short five lemma.

The guiding idea is:

$$\text{Do not chase the whole diagram at once. Identify the structure first.}$$

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1. Two-out-of-three for exact complexes

Let

$$0⟶A_{\bullet }⟶B_{\bullet }⟶C_{\bullet }⟶0$$

be a short exact sequence of complexes in an abelian category.

We claim that if two of

$$A_{\bullet },B_{\bullet },C_{\bullet }$$

are exact, then the third is exact.

Proof

A short exact sequence of complexes induces a long exact sequence in homology:

$$\cdots \to H_n(A)\to H_n(B)\to H_n(C)\stackrel{\delta }{\to }{H}_{n-1}(A)\to H_{n-1}(B)\to \cdots .$$

A complex $X_{\bullet }$$X_\bullet$ is exact if and only if

$$H_n(X)=0\text{for all }n.$$

Suppose first that $A_{\bullet }$$A_\bullet$ and $B_{\bullet }$$B_\bullet$ are exact. Then

$$H_n(A)=H_n(B)=H_{n-1}(A)=0.$$

The exact segment

$$H_n(B)\to H_n(C)\to H_{n-1}(A)$$

therefore becomes

$$0\to H_n(C)\to 0.$$

Thus $H_n(C)=0$$H_n(C)=0$ for all $n$$n$, so $C_{\bullet }$$C_\bullet$ is exact.

Suppose next that $A_{\bullet }$$A_\bullet$ and $C_{\bullet }$$C_\bullet$ are exact. The exact segment

$$H_n(A)\to H_n(B)\to H_n(C)$$

becomes

$$0\to H_n(B)\to 0.$$

Thus $H_n(B)=0$$H_n(B)=0$ for all $n$$n$, so $B_{\bullet }$$B_\bullet$ is exact.

Finally, suppose that $B_{\bullet }$$B_\bullet$ and $C_{\bullet }$$C_\bullet$ are exact. The exact segment

$$H_{n+1}(C)\to H_n(A)\to H_n(B)$$

becomes

$$0\to H_n(A)\to 0.$$

Thus $H_n(A)=0$$H_n(A)=0$ for all $n$$n$, so $A_{\bullet }$$A_\bullet$ is exact.

Therefore exactness satisfies two-out-of-three in a short exact sequence of complexes.

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2. Subobjects and quotients of three-term complexes

We first isolate a small observation.

Lemma

Let

$$X_2\stackrel{d_2^X}{\to }{X}_1\stackrel{d_1^X}{\to }{X}_0$$

be a three-term complex, so

$$d_1^X{d}_2^X=0.$$

Suppose

$$\begin{array}{ccccc}{S}_2& \stackrel{d_2^S}{\to }& S_1& \stackrel{d_1^S}{\to }& S_0\\ \downarrow i_2& & \downarrow i_1& & \downarrow i_0\\ X_2& \stackrel{d_2^X}{\to }& X_1& \stackrel{d_1^X}{\to }& X_0\end{array}$$

is a commutative diagram and $i_0,i_1,i_2$$i_0,i_1,i_2$ are monomorphisms. Then

$$S_2\stackrel{d_2^S}{\to }{S}_1\stackrel{d_1^S}{\to }{S}_0$$

is also a complex.

Dually, suppose

$$\begin{array}{ccccc}{X}_2& \stackrel{d_2^X}{\to }& X_1& \stackrel{d_1^X}{\to }& X_0\\ \downarrow q_2& & \downarrow q_1& & \downarrow q_0\\ Q_2& \stackrel{d_2^Q}{\to }& Q_1& \stackrel{d_1^Q}{\to }& Q_0\end{array}$$

is a commutative diagram and $q_0,q_1,q_2$$q_0,q_1,q_2$ are epimorphisms. Then

$$Q_2\stackrel{d_2^Q}{\to }{Q}_1\stackrel{d_1^Q}{\to }{Q}_0$$

is also a complex.

Proof

For the subobject statement, compute:

$$i_0{d}_1^S{d}_2^S=d_1^X{d}_2^X{i}_2=0.$$

Since $i_0$$i_0$ is monic,

$$d_1^S{d}_2^S=0.$$

Hence $S_{\bullet }$$S_\bullet$ is a complex.

For the quotient statement, by duality or compute:

$$d_1^Q{d}_2^Q{q}_2=q_0{d}_1^X{d}_2^X=0.$$

Since $q_2$$q_2$ is epic,

$$d_1^Q{d}_2^Q=0.$$

Hence $Q_{\bullet }$$Q_\bullet$ is a complex.

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3. The $3\times 3$$3\times3$ lemma

Suppose we have a commutative diagram in an abelian category:

$$\begin{array}{ccccccccc}0& ⟶& A^{\prime }& \stackrel{a^{\prime }}{\to }& B^{\prime }& \stackrel{b^{\prime }}{\to }& C^{\prime }& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& A& \stackrel{a}{\to }& B& \stackrel{b}{\to }& C& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& A^{″}& \stackrel{a^{″}}{\to }& B^{″}& \stackrel{b^{″}}{\to }& C^{″}& ⟶& 0\end{array}$$

and suppose every column is exact.

We want to prove:

1. if the bottom two rows are exact, then the top row is exact;

2. if the top two rows are exact, then the bottom row is exact;

3. if the top and bottom rows are exact and $ba=0$$ba=0$, then the middle row is exact.

Proof

Write the three rows as

$$R^{\prime }=(0\to A^{\prime }\stackrel{a^{\prime }}{\to }{B}^{\prime }\stackrel{b^{\prime }}{\to }{C}^{\prime }\to 0),$$

$$R=(0\to A\stackrel{a}{\to }B\stackrel{b}{\to }C\to 0),$$

and

$$R^{″}=(0\to A^{″}\stackrel{a^{″}}{\to }{B}^{″}\stackrel{b^{″}}{\to }{C}^{″}\to 0).$$

A row is exact precisely when the corresponding finite complex is exact.

Since every column is exact, once the rows involved are complexes, the vertical maps give a short exact sequence of complexes:

$$0⟶R^{\prime }⟶R⟶R^{″}⟶0.$$

Case 1: the bottom two rows are exact

Then $R$$R$ and $R^{″}$$R''$ are exact complexes. In particular, $R$$R$ is a complex.

Since every column is exact, the maps

$$A^{\prime }\to A,B^{\prime }\to B,C^{\prime }\to C$$

are monomorphisms. Hence $R^{\prime }$$R'$ is a subobject of $R$$R$ in the category of three-term diagrams.

By the previous lemma, $R^{\prime }$$R'$ is a complex.

Therefore we have a short exact sequence of complexes:

$$0⟶R^{\prime }⟶R⟶R^{″}⟶0.$$

Since $R$$R$ and $R^{″}$$R''$ are exact, two-out-of-three implies that $R^{\prime }$$R'$ is exact. Hence the top row is exact.

Case 2: the top two rows are exact

Then $R^{\prime }$$R'$ and $R$$R$ are exact complexes. In particular, $R$$R$ is a complex.

Since every column is exact, the maps

$$A\to A^{″},B\to B^{″},C\to C^{″}$$

are epimorphisms. Hence $R^{″}$$R''$ is a quotient object of $R$$R$ in the category of three-term diagrams.

By the previous lemma, $R^{″}$$R''$ is a complex.

Therefore we have a short exact sequence of complexes:

$$0⟶R^{\prime }⟶R⟶R^{″}⟶0.$$

Since $R^{\prime }$$R'$ and $R$$R$ are exact, two-out-of-three implies that $R^{″}$$R''$ is exact. Hence the bottom row is exact.

Case 3: the top and bottom rows are exact, and $ba=0$$ba=0$

The top and bottom rows being exact means that $R^{\prime }$$R'$ and $R^{″}$$R''$ are exact complexes.

The additional assumption

$$A\stackrel{a}{\to }B\stackrel{b}{\to }C=0$$

is precisely the assertion that $R$$R$ is a complex.

Thus we again have a short exact sequence of complexes:

$$0⟶R^{\prime }⟶R⟶R^{″}⟶0.$$

Since $R^{\prime }$$R'$ and $R^{″}$$R''$ are exact, two-out-of-three implies that $R$$R$ is exact. Hence the middle row is exact.

This proves the $3\times 3$$3\times3$ lemma.

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4. The short five lemma

Lemma

Let $\mathcal{A}$$\mathcal A$ be an abelian category. Suppose we have a commutative diagram whose rows are short exact:

$$\begin{array}{ccccccccc}0& \to & A^{\prime }& \stackrel{i^{\prime }}{\to }& B^{\prime }& \stackrel{p^{\prime }}{\to }& C^{\prime }& \to & 0\\ & & \downarrow f& & \downarrow g& & \downarrow h& & \\ 0& \to & A& \stackrel{i}{\to }& B& \stackrel{p}{\to }& C& \to & 0.\end{array}$$

If $f$$f$ and $h$$h$ are isomorphisms, then $g$$g$ is an isomorphism.

Proof using the arrow category

Regard the vertical maps

$$f:A^{\prime }\to A,g:B^{\prime }\to B,h:C^{\prime }\to C$$

as objects of the arrow category

$$\mathrm{Arr}(\mathcal{A})={\mathcal{A}}^{[1]}.$$

The two short exact rows say exactly that these three arrow-objects form a short exact sequence in $\mathrm{Arr}(\mathcal{A})$$\operatorname{Arr}(\mathcal A)$:

$$0⟶f⟶g⟶h⟶0.$$

Apply the kernel functor

$$\ker :\mathrm{Arr}(\mathcal{A})⟶\mathcal{A}.$$

Since $\ker$$\ker$ is left exact, we obtain an exact sequence:

$$0⟶\ker f⟶\ker g⟶\ker h.$$

Because $f$$f$ and $h$$h$ are isomorphisms,

$$\ker f=0,\ker h=0.$$

Hence

$$0⟶\ker g⟶0,$$

so

$$\ker g=0.$$

Therefore $g$$g$ is monic.

Dually, apply the cokernel functor

$$\mathrm{coker}:\mathrm{Arr}(\mathcal{A})⟶\mathcal{A}.$$

Since $\mathrm{coker}$$\operatorname{coker}$ is right exact, we obtain an exact sequence:

$$\mathrm{coker}f⟶\mathrm{coker}g⟶\mathrm{coker}h⟶0.$$

Again, because $f$$f$ and $h$$h$ are isomorphisms,

$$\mathrm{coker}f=0,\mathrm{coker}h=0.$$

Hence

$$0⟶\mathrm{coker}g⟶0,$$

so

$$\mathrm{coker}g=0.$$

Therefore $g$$g$ is epic.

In an abelian category, a morphism is an isomorphism if and only if its kernel and cokernel are both zero. Hence $g$$g$ is an isomorphism. $◻$$\square$

Actually I have another proof, see 10.1

https://www.researchgate.net/publication/405327103_Morphism_Tests_and_the_Logic_of_Separation_Property_Axiomatization_by_Uniqueness_of_Hom-Restriction

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5. A functorial comparison lemma

The reduction of the five lemma to the short five lemma rests on two small comparison facts.

Lemma: functorial comparison for cokernels and kernels

Let $\mathcal{A}$$\mathcal A$ be an abelian category.

Cokernel comparison

Suppose we have a commutative square

$$\begin{array}{ccc}{A}_1& \stackrel{a}{\to }& A_2\\ \downarrow e& & \downarrow u\\ B_1& \stackrel{b}{\to }& B_2\end{array}$$

where $e$$e$ is epic and $u$$u$ is an isomorphism. Then the morphism induced by the cokernel functor

$$\mathrm{coker}(a)⟶\mathrm{coker}(b)$$

is an isomorphism.

Kernel comparison

Dually, suppose we have a commutative square

$$\begin{array}{ccc}{A}_4& \stackrel{a}{\to }& A_5\\ \downarrow u& & \downarrow m\\ B_4& \stackrel{b}{\to }& B_5\end{array}$$

where $u$$u$ is an isomorphism and $m$$m$ is monic. Then the morphism induced by the kernel functor

$$\ker (a)⟶\ker (b)$$

is an isomorphism.

Proof

We prove the cokernel statement first.

The square says

$$ua=be.$$

Since $u$$u$ is an isomorphism, the cokernel of $ua$$ua$ is canonically identified with the cokernel of $a$$a$:

$$\mathrm{coker}(a)\cong \mathrm{coker}(ua).$$

Concretely, if

$$q_a:A_2\to \mathrm{coker}(a)$$

is the cokernel of $a$$a$, then

$$q_a{u}^{-1}:B_2\to \mathrm{coker}(a)$$

is the cokernel of $ua$$ua$.

Since $e$$e$ is epic, $be$$be$ and $b$$b$ have the same cokernel. Indeed, a morphism

$$t:B_2\to Z$$

satisfies

$$tbe=0$$

if and only if

$$tb=0,$$

because $e$$e$ is epic.

Thus

$$\mathrm{coker}(be)\cong \mathrm{coker}(b).$$

Using $ua=be$$ua=be$, we obtain

$$\mathrm{coker}(a)\cong \mathrm{coker}(ua)=\mathrm{coker}(be)\cong \mathrm{coker}(b).$$

This is exactly the morphism induced by applying

$$\mathrm{coker}:\mathrm{Arr}(\mathcal{A})\to \mathcal{A}$$

to the original square.

The kernel statement is dual.

The square says

$$ma=bu.$$

Since $u$$u$ is an isomorphism, the kernel of $bu$$bu$ is canonically identified with the kernel of $b$$b$:

$$\ker (bu)\cong \ker (b).$$

Since $m$$m$ is monic, $ma$$ma$ and $a$$a$ have the same kernel. Indeed, a morphism

$$s:T\to A_4$$

satisfies

$$mas=0$$

if and only if

$$as=0.$$

Therefore

$$\ker (a)\cong \ker (ma).$$

Using $ma=bu$$ma=bu$, we obtain

$$\ker (a)\cong \ker (ma)=\ker (bu)\cong \ker (b).$$

This is exactly the morphism induced by applying

$$\ker :\mathrm{Arr}(\mathcal{A})\to \mathcal{A}$$

to the original square.

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6. Reducing the five lemma to the short five lemma

We now prove the strong form of the five lemma.

Suppose we have a commutative diagram with exact rows:

$$\begin{array}{ccccccccc}{A}_1& \stackrel{a_1}{\to }& A_2& \stackrel{a_2}{\to }& A_3& \stackrel{a_3}{\to }& A_4& \stackrel{a_4}{\to }& A_5\\ \downarrow f_1& & \downarrow f_2& & \downarrow f_3& & \downarrow f_4& & \downarrow f_5\\ B_1& \stackrel{b_1}{\to }& B_2& \stackrel{b_2}{\to }& B_3& \stackrel{b_3}{\to }& B_4& \stackrel{b_4}{\to }& B_5.\end{array}$$

Assume

$$f_1\text{ is epic},f_2\text{ and }{f}_4\text{ are isomorphisms},f_5\text{ is monic}.$$

We prove that

$$f_3:A_3\to B_3$$

is an isomorphism.

Step 1: cut $A_3$$A_3$ into a short exact sequence

Define

$$C_A:=\mathrm{coker}(a_1),K_A:=\ker (a_4).$$

Let

$$q_A:A_2\to C_A$$

be the cokernel of $a_1$$a_1$, and let

$$k_A:K_A\to A_4$$

be the kernel of $a_4$$a_4$.

Since the top row is exact at $A_2$$A_2$,

$$\ker (a_2)=\mathrm{im}(a_1).$$

Thus $a_2$$a_2$ kills $a_1$$a_1$, so $a_2$$a_2$ factors uniquely through $q_A$$q_A$:

$$i_A:C_A\to A_3,i_A{q}_A=a_2.$$

By the first isomorphism theorem,

$$C_A=\mathrm{coker}(a_1)\cong A_2/\mathrm{im}(a_1)=A_2/\ker (a_2)\cong \mathrm{im}(a_2).$$

Hence $i_A$$i_A$ is monic and

$$\mathrm{im}(i_A)=\mathrm{im}(a_2).$$

On the other hand, exactness at $A_4$$A_4$ gives

$$\mathrm{im}(a_3)=\ker (a_4)=K_A.$$

Thus $a_3$$a_3$ factors uniquely through $k_A$$k_A$:

$$p_A:A_3\to K_A,k_A{p}_A=a_3.$$

Since

$$\mathrm{im}(a_3)=K_A,$$

the morphism $p_A$$p_A$ is epic.

Because $k_A$$k_A$ is monic,

$$\ker (p_A)=\ker (a_3).$$

Exactness at $A_3$$A_3$ gives

$$\ker (a_3)=\mathrm{im}(a_2)=\mathrm{im}(i_A).$$

Therefore we obtain a short exact sequence:

$$0⟶C_A\stackrel{i_A}{\to }{A}_3\stackrel{p_A}{\to }{K}_A⟶0.$$

Equivalently,

$$0⟶\mathrm{coker}(a_1)⟶A_3⟶\ker (a_4)⟶0.$$

Applying the same construction to the bottom row gives

$$C_B:=\mathrm{coker}(b_1),K_B:=\ker (b_4),$$

and a short exact sequence

$$0⟶C_B\stackrel{i_B}{\to }{B}_3\stackrel{p_B}{\to }{K}_B⟶0.$$

Step 2: the original diagram induces a diagram of short exact sequences

The original five-term diagram induces

$$\begin{array}{ccccccccc}0& \to & C_A& \stackrel{i_A}{\to }& A_3& \stackrel{p_A}{\to }& K_A& \to & 0\\ & & \downarrow \gamma & & \downarrow f_3& & \downarrow \delta & & \\ 0& \to & C_B& \stackrel{i_B}{\to }& B_3& \stackrel{p_B}{\to }& K_B& \to & 0.\end{array}$$

Here

$$\gamma :C_A=\mathrm{coker}(a_1)\to C_B=\mathrm{coker}(b_1)$$

is obtained by applying the cokernel functor to the left square:

$$\begin{array}{ccc}{A}_1& \stackrel{a_1}{\to }& A_2\\ \downarrow f_1& & \downarrow f_2\\ B_1& \stackrel{b_1}{\to }& B_2.\end{array}$$

Similarly,

$$\delta :K_A=\ker (a_4)\to K_B=\ker (b_4)$$

is obtained by applying the kernel functor to the right square:

$$\begin{array}{ccc}{A}_4& \stackrel{a_4}{\to }& A_5\\ \downarrow f_4& & \downarrow f_5\\ B_4& \stackrel{b_4}{\to }& B_5.\end{array}$$

The two squares in the short exact sequence diagram commute by functoriality and by the commutativity of the original diagram.

Step 3: identify the two outer vertical maps

By the functorial comparison lemma applied to the left square,

$$\gamma :\mathrm{coker}(a_1)⟶\mathrm{coker}(b_1)$$

is an isomorphism, because

$$f_1\text{ is epic}\text{and}{f}_2\text{ is an isomorphism}.$$

By the dual part of the same lemma applied to the right square,

$$\delta :\ker (a_4)⟶\ker (b_4)$$

is an isomorphism, because

$$f_4\text{ is an isomorphism}\text{and}{f}_5\text{ is monic}.$$

Thus we have a commutative diagram of short exact sequences:

$$\begin{array}{ccccccccc}0& \to & \mathrm{coker}(a_1)& \to & A_3& \to & \ker (a_4)& \to & 0\\ & & \downarrow \sim & & \downarrow f_3& & \downarrow \sim & & \\ 0& \to & \mathrm{coker}(b_1)& \to & B_3& \to & \ker (b_4)& \to & 0.\end{array}$$

By the short five lemma, the middle vertical map

$$f_3:A_3\to B_3$$

is an isomorphism.

This proves the five lemma.

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7. Summary

The five lemma is not primarily a mysterious diagram chase. Its structure is:

$$A_3\text{ is an extension of }\ker (a_4)\text{ by }\mathrm{coker}(a_1).$$

Similarly,

$$B_3\text{ is an extension of }\ker (b_4)\text{ by }\mathrm{coker}(b_1).$$

The left-side hypotheses

$$f_1\text{ epic},f_2\text{ iso}$$

ensure

$$\mathrm{coker}(a_1)\cong \mathrm{coker}(b_1).$$

The right-side hypotheses

$$f_4\text{ iso},f_5\text{ monic}$$

ensure

$$\ker (a_4)\cong \ker (b_4).$$

So the middle objects $A_3$$A_3$ and $B_3$$B_3$ are extensions with isomorphic left and right parts. The short five lemma then says that the middle map between these extensions is also an isomorphism.

In one sentence:

$$\boxed{{\displaystyle \text{five lemma = cut the middle term into a short exact sequence, then apply the short five lemma.}}}$$