Homological algebra MOC

Five lemma

If the following diagram commutes in 𝖦𝗋𝗉 with both rows exact

and 𝛾2,𝛾4 are isomorphisms, 𝛾1 epimorphism, and 𝛾5 monomorphism then 𝛾3 is an isomorphism.¹ homology

Proof

The proof involves proving the two “four lemmata”, by Diagram chasing. We will use additive notation for group operations, but the groups in question need not be abelian.

First we use the fact that 𝛾2,𝛾4 are epic and 𝛾5 is monic to show that 𝛾3 is epic.

1. Let 𝑏3 ∈𝐵3
2. By epi 𝛾4(𝑎4) =𝛽3(𝑏3) for some 𝑎4 ∈𝐴4
3. By commutativity 𝛽4𝛾4(𝑎4) =𝛾5𝛼4(𝑎4)
4. By exactness 0 =𝛽4𝛽3(𝑏3) =𝛽4𝛾4(𝑎4) =𝛾5𝛼4(𝑎4)
5. By mono 𝛼4(𝑎4) =0
6. By exactness 𝑎4 ∈ker⁡𝛼4 =im⁡𝛼3
7. Thus 𝑎4 =𝛼3(𝑎3) for some 𝑎3 ∈𝐴3
8. Thus 𝛽3𝛾3(𝑎3) =𝛾4𝛼3(𝑎3) =𝛾4(𝑎4) =𝛽3(𝑏3)
9. Thus 𝛽3(𝑏3) −𝛽3𝛾3(𝑎3) =0
10. By homo 𝛽3(𝑏3 −𝛾3(𝑎3)) =0
11. By exactness 𝑏3 −𝛾3(𝑎3) ∈ker⁡𝛽3 =im⁡𝛽2
12. Thus 𝑏3 −𝛾3(𝑎3) =𝛽2(𝑏2) for some 𝑏2 ∈𝛽2
13. By epi 𝑏2 =𝛾2(𝑎2) for some 𝑎2 ∈𝐴2
14. By commutativity 𝛽2𝛾2(𝑎2) =𝛾3𝛼2(𝑎2) =𝑏3 −𝛾3(𝑎3)
15. By homo 𝛾3(𝛼2(𝑎2) +𝑎3) =𝛾3𝛼2(𝑎2) +𝛾3(𝑎3) =𝑏3 −𝛾3(𝑎3) +𝛾3(𝑎3) =𝑏3

Therefore 𝛾3 is epic. Now we will use the fact that 𝛾2,𝛾4 are monic and 𝛾1 is epic to show that 𝛾3 is monic.

1. Let 𝑎3 ∈ker⁡𝛾3, so 𝛾3(𝑎3) =0
2. By homo 𝛽3𝛾3(𝑎3) =0
3. By commutativity 𝛾4𝛼3(𝑎3) =0
4. By mono 𝛼3(𝑎3) =0
5. By exactness 𝑎3 ∈ker⁡𝛼3 =im⁡𝛼2
6. Thus 𝑎3 =𝛼2(𝑎2) for some 𝑎2 ∈𝐴2
7. By commutativity 𝛽2𝛾2(𝑎2) =𝛾3𝛼2(𝑎2) =𝛾3(𝑎3) =0
8. By exactness 𝛾2(𝑎2) ∈ker⁡𝛽2 =im⁡𝛽1
9. Thus 𝛾2(𝑎2) =𝛽1(𝑏1) for some 𝑏1 ∈𝐵1
10. By epi 𝑏1 =𝛾1(𝑎1) for some 𝑎1 ∈𝐴1
11. By commutativity 𝛾2𝛼1(𝑎1) =𝛽1𝛾1(𝑎1) =𝛾2(𝑎2)
12. By mono 𝛼1(𝑎1) =𝑎2
13. By exactness 𝛼2𝛼1(𝑎1) =𝛼2(𝑎2) =𝑎3 =0

Therefore 𝛾3 is monic.

Every Module is a group, and every abelian category has a representation as a module category (Freyd-Mitchell theorem), so the lemma holds for module and abelian categories,

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Footnotes

1. 2010, Algebraische Topologie, ¶3.1.10, p.130ff ↩