Nine lemma

The nine lemma , also called the 3x3 lemma because of the structure of the diagram below , is a mathematical statement about commuting diagrams and exact sequences that is valid for every Abelian category as well as for the category of groups .

statement

Is (in an Abelian category or the category of groups) the diagram

commutative and if all columns and the two lower lines are exact, the upper line is also exact. The same applies: If all columns and the top two lines are exact, the bottom line is also exact.

proof

Proof is provided by chart hunting , initially assuming that the chart concerns the category of groups. For the sake of simplicity, let us denote all horizontal maps with h and all vertical maps with v . The neutral element of each group is called . The proof shows the typical characteristic of diagram hunts, that the written proof consists of nothing but trivial individual steps, which together, however, seem confusing or unmotivated - only when you follow the steps on the diagram, the connections become clear. ${\displaystyle e}$

First, let all columns and the bottom two rows be exact.

• Is with so . From this it follows with the injectivity of also and with that of finally . ${\displaystyle a_{1}\in A_{1}}$ ${\displaystyle h(a_{1})=e}$ ${\displaystyle h(v(a_{1}))=v(h(a_{1}))=v(e)=e}$ ${\displaystyle h\colon A_{2}\to B_{2}}$ ${\displaystyle v(a_{1})=0}$ ${\displaystyle v\colon A_{1}\to A_{2}}$ ${\displaystyle a_{1}=e}$
• Is , so is , so . ${\displaystyle a_{1}\in A_{1}}$ ${\displaystyle v(h(h(a_{1})))=h(h(v(a_{1})))=e}$ ${\displaystyle h(h(a_{1}))=e}$
• Is with , so , so for one . From also follows , so for a . Then is what already follows. ${\displaystyle b_{1}\in B_{1}}$ ${\displaystyle h(b_{1})=e}$ ${\displaystyle h(v(b_{1}))=v(h(b_{1}))=e}$ ${\displaystyle v(b_{1})=h(a_{2})}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle h(v(a_{2}))=v(h(a_{2}))=v(v(b_{1}))=e}$ ${\displaystyle v(a_{2})=e}$ ${\displaystyle a_{2}=v(a_{1})}$ ${\displaystyle a_{1}\in A_{1}}$ ${\displaystyle v(h(a_{1}))=h(v(a_{1}))=h(a_{2})=v(b_{1})}$ ${\displaystyle b_{1}=h(a_{1})}$
• Is so there is one with . Because there is a with . Next there is a with , well . Thus differ and in order for a suitable , ie it applies . Then and finally . ${\displaystyle c_{1}\in C_{1}}$ ${\displaystyle b_{2}\in B_{2}}$ ${\displaystyle h(b_{2})=v(c_{1})}$ ${\displaystyle h(v(b_{2}))=v(h(b_{2}))=v(v(c_{1}))=e}$ ${\displaystyle a_{3}\in A_{3}}$ ${\displaystyle h(a_{3})=v(b_{2})}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle v(a_{2})=a_{3}}$ ${\displaystyle v(h(a_{2}))=h(v(a_{2}))=h(a_{3})=v(b_{2})}$ ${\displaystyle h(a_{2})}$ ${\displaystyle b_{2}}$ ${\displaystyle v(b_{1})}$ ${\displaystyle b_{1}\in B_{1}}$ ${\displaystyle b_{2}=v(b_{1})\cdot h(a_{2})}$ ${\displaystyle v(c_{1})=h(b_{2})=h(v(b_{1})\cdot h(a_{2}))=h(v(b_{1}))\cdot h(h(a_{2}))=h(v(b_{1}))=v(h(b_{1}))}$ ${\displaystyle c_{1}=h(b_{1})}$

All points together show the accuracy of the first line.

Now all columns and the top two lines are exact.

• Is , so for one and then for one , respectively by surjectivity of and . Then is . ${\displaystyle c_{3}\in C_{3}}$ ${\displaystyle c_{3}=v(c_{2})}$ ${\displaystyle c_{2}\in C_{2}}$ ${\displaystyle c_{2}=h(b_{2})}$ ${\displaystyle b_{2}\in B_{2}}$ ${\displaystyle v\colon C_{2}\to C_{3}}$ ${\displaystyle h\colon B_{2}\to C_{2}}$ ${\displaystyle h(v(b_{2}))=v(h(b_{2}))=c_{3}}$
• Is so for one . Then . ${\displaystyle a_{3}\in A_{3}}$ ${\displaystyle a_{3}=v(a_{2})}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle h(h(a_{3}))=h(h(v(a_{2})))=v(h(h(a_{2})))=v(e)=e}$
• Is with and we choose one with , so , so for one . Next for one . Then so for one . Finally is . ${\displaystyle b_{3}\in B_{3}}$ ${\displaystyle h(b_{3})=e}$ ${\displaystyle b_{2}\in B_{2}}$ ${\displaystyle v(b_{2})=b_{3}}$ ${\displaystyle v(h(b_{2}))=h(v(b_{2}))=h(b_{3})=e}$ ${\displaystyle h(b_{2})=v(c_{1})}$ ${\displaystyle c_{1}\in C_{1}}$ ${\displaystyle c_{1}=h(b_{1})}$ ${\displaystyle b_{1}\in B_{1}}$ ${\displaystyle h(v(b_{1}))=v(h(b_{1}))=v(c_{1})=h(b_{2})}$ ${\displaystyle b_{2}=v(b_{1})\cdot h(a_{2})}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle h(v(a_{2}))=v(h(a_{2}))=v(v(b_{1}))\cdot v(h(a_{2}))=v(v(b_{1})\cdot h(a_{2}))=v(b_{2})=b_{3}}$
• Is with and we vote with , so , so for one . It is , therefore, already . Hence for one . From already follows and thus . ${\displaystyle a_{3}\in A_{3}}$ ${\displaystyle h(a_{3})=e}$ ${\displaystyle a_{2}\in A_{2}}$ ${\displaystyle v(a_{2})=a_{3}}$ ${\displaystyle v(h(a_{2}))=h(v(a_{2}))=h(a_{3})=e}$ ${\displaystyle h(a_{2})=v(b_{1})}$ ${\displaystyle b_{1}\in B_{1}}$ ${\displaystyle v(h(b_{1}))=h(v(b_{1}))=h(h(a_{2}))=e}$ ${\displaystyle h(b_{1})=e}$ ${\displaystyle b_{1}=h(a_{1})}$ ${\displaystyle a_{1}\in A_{1}}$ ${\displaystyle h(v(a_{1}))=v(h(a_{1}))=v(b_{1})=h(a_{2})}$ ${\displaystyle a_{2}=v(a_{1})}$ ${\displaystyle a_{3}=v(a_{2})=v(v(a_{1}))=e}$

Together this gives the accuracy of the last line.

The proof initially carried out for groups also applies (if necessary translated into additive notation) for Abelian groups or for modules over a ring . Thanks to Mitchell's embedding theorem , this is already sufficient to prove the nine lemma for all Abelian categories.

See also

• Five lemma
• Snake lemma

Individual evidence

1. ↑ Saunders Mac Lane : Homology , Springer Grundlehren der Mathematischen Wissenschaften Volume 114 (1967), Chapter II, Lemma 5.1 (The 3x3 Lemma)