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Home , Commutative diagram, Exact sequence, Snake lemma, Splitting lemma

DIAGRAM CHASES: EXAMPLES & EXERCISES

PETER J. HAINE

1. Examples of Diagram Chases

1.1. Notation. Throughout, we write 푅 for a commutative and Mod푅 for the of 푅-modules. 1.2. Recollection.

• Let 푘 be a . Then Mod푘 is the category Vect푘 of 푘-vector spaces and 푘-linear transformations. • The category Mod퐙 of modules over the ring of integers 퐙 is simply the category Ab of abelian groups. 1.3. Lemma (5-lemma). Consider a

훼1 훼2 훼3 훼4 푀1 푀2 푀3 푀4 푀5 (1.4) 훾1 훾2 훾3 훾4 훾5 ′ ′ ′ ′ ′ 푀1 ′ 푀2 ′ 푀3 ′ 푀4 ′ 푀5 . 훼1 훼2 훼3 훼4 in Mod푅. If the rows are exact, then the following statements hold.

(1.3.a) If 훾2 and 훾4 are injective and 훾1 is surjective, then 훾3 is injective. (1.3.b) If 훾2 and 훾4 are surjective and 훾5 is injective, then 훾3 is surjective.

Proof. First we prove (1.3.a). So suppose that 훾2 and 훾4 are injective and 훾1 is surjective. Start with 푚3 ∈ 푀3 with the property that 훾3(푚3) = 0. The goal is to show that 푚3 = 0. ′ Then since 훾3(푚3) = 0, we also know that 훼3훾3(푚3) = 0, hence by the commutativity of the diagram (2.5) we have 훾4훼3(푚3) = 0. Since 훾4 is injective, this implies that 훼3(푚3) = 0, hence 푚3 ∈ ker 훼3. Since the top row is exact, ker 훼3 = im 훼2, so there exists an 푚2 ∈ 푀2 such that 훼2(푚2) = 푚3. Then since 훾3(푚3) = 0 we see that 훾3훼2(푚2) = 0. Then by the ′ ′ commutativity of the diagram (1.4) we see that 훼2훾2(푚2) = 0. Therefore 훾2(푚2) ∈ ker 훼2. ′ ′ ′ ′ Since the bottom row is exact, ker(훼2) = im(훼1), so there exists an 푚1 ∈ 푀1 so that ′ ′ ′ 훼1(푚1) = 훾2(푚2). Since 훾1 is surjective, there exists a 푚1 ∈ 푀1 so that 훾1(푚1) = 푚1. Then by the commutativity of the diagram (1.4), we have

′ ′ ′ 훾2훼1(푚1) = 훼1훾1(푚1) = 훼1(푚1) = 훾2(푚2).

Thus by the injectivity of 훾2 we have 훼1(푚1) = 푚2. This shows that

훼2훼1(푚1) = 훼2(푚2) = 푚3.

Date: January 9, 2018. 1 2 PETER J. HAINE

But by the exactness of the top row, we have 훼2훼1(푚1) = 0, so 푚3 = 0, as desired. Hence ker 훾3 = 0, so 훾3 is injective. A summary of this chase is displayed below.

푚1 푚2 푚3 훼3(푚3)

′ 푚1 훾2(푚2) 0 0 . † Now we prove (1.3.b). So suppose that 훾2 and 훾4 are surjective and 훾5 is injective. Start ′ ′ ′ by considering an element 푚3 ∈ 푀3: the goal is to show that 푚3 lies in the of 훾3. ′ ′ ′ Map 푚3 to 훼3(푚3). Then by the surjectivity of 훾4 there exists an element 푚4 ∈ 푀4 so that ′ ′ 훾4(푚4) = 훼3(푚3). Now consider 훼4(푚4). Notice that by the commutativity of the diagram (1.4), we have ′ ′ ′ 훾5훼4(푚4) = 훼4훾4(푚4) = 훼4훼3(푚3). ′ ′ Then by the exactness of the bottom row, 훼4훼3 = 0, so 훾5훼4(푚4) = 0. Therefore, by the injectivity of 훾5 we have 훼4(푚4) = 0, so 푚4 ∈ ker(훼4). By the exactness of the top row, ker(훼4) = im(훼3), so there exists 푚3 ∈ 푀3 so that 훼3(푚3) = 푚4. ′ Now consider 푚3 − 훾3(푚3). Notice that by the commutativity of the diagram (1.4), ′ ′ ′ ′ ′ ′ ′ 훼3(푚3 − 훾3(푚3)) = 훼3(푚3) − 훼3훾3(푚3) = 훼3(푚3) − 훾4훼3(푚3) ′ ′ ′ ′ ′ ′ = 훼3(푚3) − 훾4(푚4) = 훼3(푚3) − 훼3(푚3) = 0. ′ ′ ′ This shows that 푚3 − 훾3(푚3) is in the of 훼3, which is equal to im 훼2 by the exactness ′ ′ ′ ′ of the lower . Hence there exists an 푚2 so that 훼2(푚2) = 푚3 − 훾3(푚3). Then by the ′ surjectivity of 훾2 there exists an 푚2 ∈ 푀2 so that 훾2(푚2) = 푚2. Now consider 푚3 + 훼2(푚2) and notice that by the commutativity of the diagram (1.4), we have ′ 훾3(푚3 + 훼2(푚2)) = 훾3(푚3) + 훾3훼2(푚2) = 훾3(푚3) + 훼2훾2(푚3) ′ ′ = 훾3(푚3) + (푚3 − 훾3(푚3)) = 푚3. □ This shows that 훾3 is surjective, as desired. 1.5. Remark. The standard application of the 5-lemma in is the fol- lowing: in the diagram (1.4) the homomorphisms 훾1, 훾2, 훾4, and 훾5 are , from which we conclude that 훾3 is an . Now we prepare to prove the (Proposition 1.14) with a few easy lemmas. First we recall a few definitions. 1.6. Recollection. Let 푅 be a commutative ring and 푓∶ 퐴 → 퐵 a of 푅-modules. Recall that the kernel of 푓 is an 푅- ker(푓) equipped with a morphism a morphism 푖∶ ker(푓) → 퐴 satisfying the following : the composite 푓푖 is the zero mor- phism and for any morphism 푔∶ 퐾 → 퐴 such that 푓푔 = 0, there exists a unique morphism ̄푔∶ 퐾 → ker(푓) making the triangle 퐾 푔 ̄푔

ker(푓) 퐴 푖 commute.

† Actually this follows immediately by duality since Mod푅 is an and could have been proven in the setting of abelian categories — maybe we’ll talk about these at some point. DIAGRAM CHASES: EXAMPLES & EXERCISES 3

The kernel may explicitly be realized as the submodule of 퐴 defined by ker(푓) ≔ { 푎 ∈ 퐴 | 푓(푎) = 0 } , so that the morphism 푖∶ ker(푓) → 퐴 in the universal property is simply the inclusion ker(푓) ↪ 퐴. 1.7. Remark. This latter description is the description we encounter in a first course in algebra. 1.8. Recollection. Recall that the of a morphism of 푅-modules 푓∶ 퐴 → 퐵 is an 푅- module coker(푓) equipped with a morphism a morphism 푞∶ 퐵 → coker(푓) satisfying the following universal property: the composite 푞푓 is the zero morphism and for any morphism ℎ∶ 퐵 → 퐶 such that ℎ푓 = 0, there exists a unique morphism ℎ∶̄ coker(푓) → 퐶 making the triangle 퐵 ℎ 푞

coker(푓) 퐴 ℎ̄ commute. The cokernel may explicitly be realized as the quotient of 퐵 by the image of 푓: coker(푓) ≔ 퐵/ im(푓) , so that the morphism 푞∶ 퐵 → coker(푓) in the universal property is simply the quotient map 퐵 ↠ coker(푓). 1.9. Remark. The cokernel is often not introduced in a first course in algebra, which iswhy we introduce it here. The “explicit” description of the cokernel is what is often introduced in commutative algebra courses, but it makes the duality between the kernel and cokernel less obvious than the universal property description. 1.10. Lemma. Let 푅 be a commutative ring and

푓 퐴 퐵

(1.11) 훼 훽

퐴′ 퐵′ 푓′ a commutative square in Mod푅.

(1.10.a) The restriction 푓|ker(훼) ∶ ker(훼) → ker(훽) is the unique 푅- making the square 푓| ker(훼) ker(훼) ker(훽)

푖훼 푖훽 퐴 퐵 푓

commute, where 푖훼 and 푖훽 denote the inclusions. Moreover, if 푓 is an injection, then so is 푓|ker(훼). 4 PETER J. HAINE

(1.10.b) Dually, there exists a unique 푅-module homomorphism 푓∶̄ coker(훼) → coker(훽) making the square

푓′ 퐴′ 퐵′

푞훼 푞훽 coker(훼) coker(훽) 푓′̄

commute, where 푞훼 and 푞훽 denote the quotient maps. Moreover, if 푓 is a surjection, then so is 푓′̄ . Proof. Since (1.10.b) is dual to (1.10.a), it suffices to prove (1.10.a). This follows easily from the universal property of the kernel. Consider the composite 훽푓푖훼. By the commutativity of the square (1.11) and the fact that 푖훼 is the inclusion of the kernel of 훼 we have that ′ 훽푓푖훼 = 푓 훼푖훼 = 0 . Thus by the universal property of ker(훽), there exists a unique 푅-module homomorphism ker(훼) → ker(훽) making the square ker(훼) ker(훽)

푖훼 푖훽 퐴 퐵 푓 commute. The fact that the unique morphism ker(훼) → ker(훽) is the restriction 푓|ker(훼) follows immediately from the definition of the restriction. The last thing to check is that if 푓 is an injection, then so is 푓|ker(훼). To see this, notice that if 푓 is an injection, then since 푖훼 is an inclusion, 푓푖훼 is also an injection, so the composite □ 푖훽 ∘ 푓|ker(훼) is an injection, which implies that 푓|ker(훼) is an injection. 1.12. Lemma. Let 푅 be a commutative ring and consider the following diagram of 푅-modules

푓 푔 퐴 퐵 퐶

(1.13) 훼 훽 훾

퐴′ 퐵′ 퐶′ . 푓′ 푔′ (1.12.a) If the top row of (1.13) is exact and 푓′ is an injection, then the sequence

푓| 푔| ker(훼) ker(훼) ker(훽) ker(훽) ker(훾) is exact. (1.12.b) Dually, if the bottom row of (1.13) is exact and 푔 is a surjection, then the sequence

푓′̄ ′̄푔 coker(훼) coker(훽) coker(훾) is exact. Proof. As usual, since (1.12.b) is dual to (1.12.a), it suffices to prove (1.12.a). The only place where we need to show exactness is at ker(훽). First let us show that im(푓|ker(훼)) ⊂ DIAGRAM CHASES: EXAMPLES & EXERCISES 5 ker(푔|ker(훽)). So suppose that 푏 ∈ ker(훽) and 푏 = 푓(푎) for some 푎 ∈ ker(훼). Then since the top row of (1.13) we see that

푔|ker(훽)(푏) = 푔(푏) = 푔푓(푎) = 0 .

Hence 푏 ∈ ker(푔|ker(훽)), so we see that im(푓|ker(훼)) ⊂ ker(푔|ker(훽)). Now let us show that ker(푔|ker(훽)) ⊂ im(푓|ker(훼)). Suppose that 푏 ∈ ker(푔|ker(훽)). Then by the exactness of the top row of (1.13) we know that 푏 = 푓(푎) for some 푎 ∈ 퐴. Hence it suffices to show that 푎 ∈ ker(훼). Since 푏 ∈ ker(훽) we know that 훽(푏) = 0. But by the commutativity of (1.13) we see that 훽(푏) = 훽푓(푎) = 푓′훼(푎) . Since 푓′ is injective, we see that 훼(푎) = 0, as desired. □

1.14. Proposition (Snake Lemma). Let 푅 be a commutative ring and consider the following diagram of 푅-modules, where the rows are exact

푓 푔 퐴 퐵 퐶 0

(1.15) 훼 훽 훾

0 퐴′ 퐵′ 퐶′ . 푓′ 푔′ There exists a connecting homomorphism 휕∶ ker(훾) → coker(훼), and the sequence

푓| 푔| 휕 푓′̄ ′̄푔 ker(훼) ker(훼) ker(훽) ker(훽) ker(훾) coker(훼) coker(훽) coker(훾)

′ ′ is exact. Moreover, if 푓 is injective, then so is 푓|ker(훼). Dually, if 푔 is surjective, then so is ̄푔 . Proof. There are three things to do: construct the connecting homomorphism, show that the connecting homomorphism is well-defined, and, in light of Lemma 1.12, show that the resulting sequence is exact at ker(훾) and coker(훼). First we construct the connecting homo- morphism. We start with an element 푐 ∈ ker(훾). Since 푔 is surjective, we can choose an element 푏 ∈ 퐵 such that 푔(푏) = 푐. By the commutativity of the right-hand square of (1.15) and the fact that 푐 ∈ ker(훾), 푔′훽(푏) = 훾푔(푏) = 훾(푐) = 0 , so by the exactness of the lower sequence of (1.15) and the injectivity of 푓′, there exists a unique 푎′ ∈ 퐴′ with the property that 푓′(푎′) = 훽(푏). We define the connecting homomor- phism 휕∶ ker(훾) → coker(훼) by the assignment 푐 ↦ [푎′] , where [푎′] denotes the class of 푎′ in coker(훼). Now let us show that 휕 is well-defined as a set-theoretic map. Suppose that 푏̃ ∈ 퐵 is another element such that 푔(푏)̃ = 푐, and let ′̃푎 ∈ 퐴′ be the unique element with the property that 푓′(′ ̃푎 ) = 훽(푏)̃ . Then since 푔 is an 푅-module homomorphism, 푔(푏 − 푏)̃ = 푔(푏) − 푔(푏)̃ = 푐 − 푐 = 0 , so by the exactness of the top row of (1.15) there exists an element 푎 ∈ 퐴 with the property that 푓(푎) = 푏 − 푏̃. By the commutativity of the left-hand square of (1.15), we see that 푓′훼(푎) = 훽푓(푎) = 훽(푏 − 푏)̃ = 훽(푏) − 훽(푏)̃ . 6 PETER J. HAINE

Since 푓′(푎′ −′ ̃푎 ) = 훽(푏) − 훽(푏)̃ and 푓′ is injective, we see that 훼(푎) = 푎′ −′ ̃푎 , that is 푎′ =′ ̃푎 + 훼(푎). Thus in coker(훼) we see that [푎′] = [′ ̃푎 + 훼(푎)] = [′ ̃푎 ] , so the connecting homomorphism is actually well-defined. Now we need to prove that 휕 is an 푅-module homomorphism, but this is easy since we now know that the value of 휕(푐) is independent of any choices we make. Suppose that we are given 푟 ∈ 푅 and 푐 ∈ ker(훾). Choose 푏 ∈ 퐵 so that 훽(푏) = 푐. Then since 훽 is an 푅- module homomorphism, we have that 훽(푟푏) = 푟푐. Let 푎′ ∈ 퐴′ be the unique element so that 푓′(푎′) = 훽(푏′). Then since 푓′ and 훽 are 푅-module homorphisms 푓′(푟푎′) = 푟푓′(푎′) = 푟훽(푏) = 훽(푟푏) , hence 휕(푟푐) = [푟푎′] = 푟[푎′] = 푟휕(푐) . ′ Similarly, suppose that 푐1, 푐2 ∈ ker(훾), and let 푏1, 푏2 ∈ 퐵 be elements such that 푔(푏1) = 푐1 ′ ′ ′ ′ ′ and 푔(푏2) = 푐2, and let 푎1, 푎2 ∈ 퐴 be the unique elements such that 푓 (푎1) = 훽(푏1) and ′ ′ ′ 푓 (푎2) = 훽(푏2). Then since 훽 and 푓 are 푅-module homomorphisms ′ ′ ′ ′ ′ ′ ′ 푓 (푎1 + 푎2) = 푓 (푎1) + 푓 (푎2) = 훽(푏1) + 훽(푏2) = 훽(푏1 + 푏2) , so ′ ′ ′ ′ 휕(푐1 + 푐2) = [푎1 + 푎2] = [푎1] + [푎2] , which completes the proof that 휕 is actually an 푅-module homomorphism. Finally, let us show that the sequence

푓| 푔| 휕 푓′̄ ′̄푔 (1.16) ker(훼) ker(훼) ker(훽) ker(훽) ker(훾) coker(훼) coker(훽) coker(훾) is exact. By Lemma 1.12 the sequence (1.16) is exact at ker(훽) and coker(훽), so we just need to show that (1.16) is exact at ker(훾) and coker(훼). First let us show that (1.16) is exact at ker(훾). Suppose that 푐 ∈ im(푔|ker(훽)) so that 푐 = 푔(푏) for some 푏 ∈ ker(훽). Then 훽(푏) = 0, so by the definition of 휕 we have that 휕(푐) = 0, hence ker(휕) ⊂ im(푔|ker(훽)). On the other hand suppose that 푐 ∈ ker(휕). Let 푏 ∈ 퐵 be such that 푔(푏) = 푐 and 푎′ ∈ 퐴′ be the unique element so that 푓′(푎′) = 훽(푏). Then by the definition of 휕 and the fact that 휕(푐) = 0, we have that [푎′] = 0, that is 푎′ ∈ im(훼), so there exists some element 푎 ∈ 퐴 so that 훼(푎) = 푎′. Thus 푓′(푎′) = 푓′훼(푎) = 훽푓(푎) . But 훽(푏) = 푓′(푎′), so we have that 훽(푏) = 훽푓(푎). Thus 푏 − 푓(푎) ∈ ker(훽). Since 푐 = 푔(푏) we have that 푐 = 푔(푏 − 푓(푎) + 푓(푎)) = 푔(푏 − 푓(푎)) + 푔푓(푎) = 푔(푏 − 푓(푎)) .

Since 푏 − 푓(푎) ∈ ker(훽), this shows that 푐 ∈ im(푔|ker(훽)), as desired. Hence the sequence (1.16) is exact at ker(훾). Now let us show that the sequence (1.16) is exact at coker(훾). Suppose that [푎′] ∈ im(휕). Then [푎′] = 휕(푐) for some 푐 ∈ ker(훾). We know that 푓′̄ [푎′] = [푓′(푎′)] . By the definition of 휕, we have that 푓′(푎′) = 훽(푏), where 푔(푏) = 푐. Hence, in coker(훽) 푓′̄ [푎′] = [푓′(푎′)] = [훽(푏)] = 0 , so [푎′] ∈ ker(푓′̄ ), as desired. DIAGRAM CHASES: EXAMPLES & EXERCISES 7

On the other hand, suppose that 푎′ ∈ 퐴′ and 푓′̄ [푎′] = 0. This means that 푓′(푎′) ∈ im(훽), so 푓′(푎′) = 훽(푏) for some 푏 ∈ 퐵. Then

훾푔(푏) = 푔′훽(푏) = 푔′푓′(푎) = 0 , by the commutativity of (1.15) and the exactness of the bottom row. Thus 푔(푏) ∈ ker(훾), and by the definition of 휕 we have that 휕푔(푏) = [푎′], so [푎′] ∈ im(휕), as desired. ′ The last statement, that if 푓 is injective, then so is 푓|ker(훼), and if 푔 is surjective, then so is ′̄푔 is just an application of Lemma 1.10. □

2. Exercises in diagram chasing The following lemmas are good exercises in diagram chasing and are both standard re- sults from homological algebra. As an exercise, try to prove these results.

2.1. Exercise. Prove (1.10.b) directly (i.e., not appealing to duality).

2.2. Lemma (). Let 푅 be a commutative ring and

푖 푝 0 푋 푍 푌 0 be a short of 푅-modules. The following are equivalent. (2.2.a) The image of 푖 is a direct summand of 푍. (2.2.b) There exists a retraction 푟 of 푖. (2.2.c) There exists a section 푠 of 푝. In this case, we say that the sequence splits, and 푍 ≅ 푋 ⊕ 푌 in such a way that 푖 is the inclusion and 푝 is the projection.

2.3. Lemma. Let 푅 be a commutative ring. Then: (2.3.a) Given an exact sequence of 푅-modules

푖 푝 0 푋 푍 푌 ,

we have ker(푝) ≅ 푋. (2.3.b) Given an exact sequence of 푅-modules

푖 푝 푋 푍 푌 0 ,

we have coker(푖) ≅ 푌. (Moreover, these isomorphisms are natural in of exact .)

The Snake lemma might come in handy to prove the next lemma. 8 PETER J. HAINE

2.4. Lemma (3 × 3 lemma). Consider a commutative diagram 0 0 0

′ 훾1 휌1 ″ 0 푀1 푀1 푀1 0

훼′ 훼 훼″

(2.5) ′ ″ 0 푀2 훾2 푀2 휌2 푀2 0

훽′ 훽 훽″

′ ″ 0 푀3 푀3 푀3 0 훾3 휌3

0 0 0 in Mod푅, where all of the columns are exact and the middle row is exact. Then the first row of (2.5) is exact if and only if the third is. Proof. First, assume that the first row of (2.5) is exact. Then by the Snake Lemma, the first two rows induce an exact sequence

휙 휓 휕 휙′ 휓′ 0 ker 훼′ ker 훼 ker 훼″ coker 훼′ coker 훼 coker 훼″ 0 . Since the columns of (2.5) are exact, 훼, 훼′, and 훼″ are injective, so the kernels of each of these morphisms are zero. Again since the columns of (2.5) are exact, by Lemma 2.3 we ′ ′ ″ ″ have 푀3 = coker 훼 , 푀3 = coker 훼, and 푀3 = coker 훼 . Hence the sequence

′ ′ ′ 휙 휓 ″ 0 푀3 푀3 푀3 0 is exact. By the definitions of 휙′ and 휓′ in the Snake Lemma and the commutativity of the ′ ′ ′ ′ ′ diagram (2.5), we see that 휙 훽 = 훾3훽 and 휓 훽 = 휌3훽. Since 훽 and 훽 are in ′ ′ Mod푅, this implies that 휙 = 훾3 and 휓 = 휌3 so the last row of (2.5) is exact. Now, assume that the last row of (2.5) is exact. The fact that the first row is exact follows by duality — we provide a seperate proof anyhow. Then by the Snake Lemma, the last two rows induce an exact sequence

휙 휓 휕 휙′ 휓′ 0 ker 훽′ ker 훽 ker 훽″ coker 훽′ coker 훽 coker 훽″ 0 . Since the columns of (2.5) are exact, 훽, 훽′, and 훽″ are surjective, so the of each of these morphisms are zero. Again since the columns of (2.5) are exact, by Lemma 2.3 we ′ ′ ″ ″ have 푀1 = ker 훽 , 푀1 = ker 훽, and 푀1 = ker 훽 . Hence the sequence

′ 휙 휓 ″ 0 푀1 푀1 푀1 0 is exact. By the definitions of 휙 and 휓 in the Snake Lemma and the commutativity of the ″ ″ ″ diagram (2.5), we see that 훼휙 = 훼훾1 and 훼 휓 = 훼 휌1. Since 훼 and 훼 are □ in Mod푅, this implies that 휙 = 훾1 and 휓 = 휌1 so the first row of (2.5) is exact.