[[Homological algebra MOC]]
# Five lemma

If the following diagram commutes in $\Grp$ with both rows [[Exact sequence|exact]]

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and $\gamma_{2},\gamma_{4}$ are isomorphisms, $\gamma_{1}$ [[Group epimorphism|epimorphism]], and $\gamma_{5}$ [[Group monomorphism|monomorphism]]
then $\gamma_{3}$ is an isomorphism.[^loose] #m/thm/homology

> [!check]- Proof
> The proof involves proving the two “four lemmas”, 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 $\gamma_{2},\gamma_{4}$ are epic and $\gamma_{5}$ is monic to show that $\gamma_{3}$ is epic.
> 
> 1. Let $b_{3} \in B_{3}$
> 2. By epi $\gamma_{4}(a_{4}) = \beta_3(b_{3})$ for some $a_{4} \in A_{4}$
> 3. By commutativity $\beta_{4}\gamma_{4}(a_{4}) = \gamma_{5}\alpha_{4}(a_{4})$
> 4. By exactness $0 = \beta_{4}\beta_{3}(b_{3}) = \beta_{4}\gamma_{4}(a_{4}) = \gamma_{5}\alpha_{4}(a_{4})$
> 5. By mono $\alpha_{4}(a_{4}) = 0$
> 6. By exactness $a_{4} \in \ker\alpha_{4}= \im \alpha_{3}$
> 7. Thus $a_{4} = \alpha_{3}(a_{3})$ for some $a_{3} \in A_{3}$
> 8. Thus $\beta_{3}\gamma_{3}(a_{3}) = \gamma_{4}\alpha_{3}(a_{3}) = \gamma_{4} (a_{4}) = \beta_{3}(b_{3})$
> 9. Thus $\beta_{3}(b_{3}) - \beta_{3}\gamma_{3}(a_{3}) = 0$
> 10. By homo $\beta_{3}(b_{3} - \gamma_{3}(a_{3})) = 0$
> 11. By exactness $b_{3} - \gamma_{3}(a_{3}) \in \ker \beta_{3} = \im \beta_{2}$
> 12. Thus $b_{3}-\gamma_{3}(a_{3})= \beta_{2}(b_{2})$ for some $b_{2}\in \beta_{2}$
> 13. By epi $b_{2} = \gamma_{2}(a_{2})$ for some $a_{2} \in A_{2}$
> 14. By commutativity $\beta_{2}\gamma_{2}(a_{2})= \gamma_{3}\alpha_{2}(a_{2}) = b_{3} - \gamma_{3}(a_{3})$ 
> 15. By homo $\gamma_{3}(\alpha_{2}(a_{2}) + a_{3}) = \gamma_{3}\alpha_{2}(a_{2})+\gamma_{3}(a_{3}) = b_{3} - \gamma_{3}(a_{3}) + \gamma_{3}(a_{3}) = b_{3}$
> 
> Therefore $\gamma_{3}$ is epic.
> Now we will use the fact that $\gamma_{2},\gamma_{4}$ are monic and $\gamma_{1}$ is epic to show that $\gamma_{3}$ is monic.
> 
> 1. Let $a_{3} \in \ker \gamma_{3}$, so $\gamma_{3}(a_{3})=0$
> 2. By homo $\beta_{3}\gamma_{3}(a_{3})=0$
> 3. By commutativity $\gamma_{4}\alpha_{3}(a_{3})=0$
> 4. By mono $\alpha_{3}(a_{3}) = 0$
> 5. By exactness $a_{3} \in \ker \alpha_{3} = \im \alpha_{2}$
> 6. Thus $a_{3} = \alpha_{2}(a_{2})$ for some $a_{2} \in A_{2}$
> 7. By commutativity $\beta_{2}\gamma_{2}(a_{2}) = \gamma_{3}\alpha_{2}(a_{2}) = \gamma_{3}(a_{3}) = 0$
> 8. By exactness $\gamma_{2}(a_{2}) \in \ker \beta_{2} = \im \beta_{1}$
> 9. Thus $\gamma_{2}(a_{2})= \beta_{1}(b_{1})$ for some $b_{1} \in B_{1}$
> 10. By epi $b_{1} = \gamma_{1}(a_{1})$ for some $a_{1} \in A_{1}$
> 11. By commutativity $\gamma_{2}\alpha_{1}(a_{1})=\beta_{1}\gamma_{1}(a_{1}) = \gamma_{2}(a_{2})$
> 12. By mono $\alpha_{1}(a_{1}) = a_{2}$
> 13. By exactness $\alpha_{2}\alpha_{1}(a_{1})= \alpha_{2}(a_{2}) = a_{3} = 0$
> 
> Therefore $\gamma_{3}$ is monic.
> <span class="QED"/>

[^loose]: 2010, [[@looseAlgebraischeTopologie2010|Algebraische Topologie]], ¶3.1.10, p.130ff

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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