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Diagram Chases: Examples & Exercises

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Diagram Chases: Examples & Exercises DIAGRAM CHASES: EXAMPLES & EXERCISES PETER J. HAINE 1. Examples of Diagram Chases 1.1. Notation. Throughout, we write 푅 for a commutative ring and Mod푅 for the category of 푅-modules. 1.2. Recollection. • Let 푘 be a field. 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 commutative diagram 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 image 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 kernel of 훼3, which is equal to im 훼2 by the exactness ′ ′ ′ ′ of the lower sequence. 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 homological algebra is the fol- lowing: in the diagram (1.4) the homomorphisms 훾1, 훾2, 훾4, and 훾5 are isomorphisms, from which we conclude that 훾3 is an isomorphism. Now we prepare to prove the Snake lemma (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 morphism of 푅-modules. Recall that the kernel of 푓 is an 푅-module ker(푓) equipped with a morphism a morphism 푖∶ ker(푓) → 퐴 satisfying the following universal property: 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 abelian category 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 cokernel 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-module homomorphism 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)).
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