Isbell's zigzag theorem

Isbell's zigzag theorem, a theorem of abstract algebra characterizing the notion of a dominion, was introduced by American mathematician John R. Isbell in 1966.^[1] Dominion is a concept in semigroup theory, within the study of the properties of epimorphisms. For example, let U is a subsemigroup of S containing U, the inclusion map ${\displaystyle U\hookrightarrow S}$ is an epimorphism if and only if ${\displaystyle {\rm {{Dom}_{S}(U)=S}}}$, furthermore, a map ${\displaystyle \alpha \colon S\to T}$ is an epimorphism if and only if ${\displaystyle {\rm {{Dom}_{T}({\rm {{im}\;\alpha )=T}}}}}$.^[2] The categories of rings and semigroups are examples of categories with non-surjective epimorphism, and the Zig-zag theorem gives necessary and sufficient conditions for determining whether or not a given morphism is epi.^[3] Proofs of this theorem are topological in nature, beginning with Isbell (1966) for semigroups, and continuing by Philip (1974), completing Isbell's original proof.^[3][4][5] The pure algebraic proofs were given by Howie (1976) and Storrer (1976).^{[3][4][note 1]}

Statement

Zig-zag

The dashed line is the spine of the zig-zag.

Zig-zag:^{[7][2][8][9][10][note 2]} If U is a submonoid of a monoid (or a subsemigroup of a semigroup) S, then a system of equalities;

${\displaystyle {\begin{aligned}d&=x_{1}u_{1},&u_{1}&=v_{1}y_{1}\\x_{i-1}v_{i-1}&=x_{i}u_{i},&u_{i}y_{i-1}&=v_{i}y_{i}\;(i=2,\dots ,m)\\x_{m}v_{m}&=u_{m+1},&u_{m+1}y_{m}&=d\end{aligned}}}$

in which ${\displaystyle u_{1},\dots ,u_{m+1},v_{1},\dots ,v_{m}\in U}$ and ${\displaystyle x_{1},\dots ,x_{m},y_{1},\dots ,y_{m}\in S}$, is called a zig-zag of length m in S over U with value d. By the spine of the zig-zag we mean the ordered (2m + 1)-tuple ${\displaystyle (u_{1},v_{1},u_{2},v_{2},\dots ,u_{m},v_{m},u_{m+1})}$.

Dominion

Dominion:^[5][6] Let U be a submonoid of a monoid (or a subsemigroup of a semigroup) S. The dominion ${\displaystyle {\rm {{Dom}_{S}(U)}}}$ is the set of all elements ${\displaystyle s\in S}$ such that, for all homomorphisms ${\displaystyle f,g:S\to T}$ coinciding on U, ${\displaystyle f(s)=g(s)}$.

We call a subsemigroup U of a semigroup U closed if ${\displaystyle {\rm {{Dom}_{S}(U)=U}}}$, and dense if ${\displaystyle {\rm {{Dom}_{S}(U)=S}}}$.^[2][12]

Isbell's zigzag theorem

Isbell's zigzag theorem:^[13]

If U is a submonoid of a monoid S then ${\displaystyle d\in {\rm {{Dom}_{S}(U)}}}$ if and only if either ${\displaystyle d\in U}$ or there exists a zig-zag in S over U with value d that is, there is a sequence of factorizations of d of the form

${\displaystyle d=x_{1}u_{1}=x_{1}v_{1}y_{1}=x_{2}u_{2}y_{1}=x_{2}v_{2}y_{2}=\cdots =x_{m}v_{m}y_{m}=u_{m+1}y_{m}}$

This statement also holds for semigroups.^{[7][14][9][4][10]}

For monoids, this theorem can be written more concisely:^[15][2][16]

Let S be a monoid, let U be a submonoid of S, and let ${\displaystyle d\in S}$. Then ${\displaystyle d\in \mathrm {Dom} _{S}(U)}$ if and only if ${\displaystyle d\otimes 1=1\otimes d}$ in the tensor product ${\displaystyle S\otimes _{U}S}$.

Application

• Let U be a commutative subsemigroup of a semigroup S. Then ${\displaystyle {\rm {{Dom}_{S}(U)}}}$ is commutative.^[10]
• Every epimorphism ${\displaystyle \alpha \colon S\to T}$ from a finite commutative semigroup S to another semigroup T is surjective.^[10]
• Inverse semigroups are absolutely closed.^[7]
• Example of non-surjective epimorphism in the category of rings:^[3] The inclusion ${\displaystyle i:(\mathbb {Z} ,\cdot )\hookrightarrow (\mathbb {Q} ,\cdot )}$ is an epimorphism in the category of all rings and ring homomorphisms by proving that any pair of ring homomorphisms ${\displaystyle \beta ,\gamma$ :\mathbb {Q} \to \mathbb {R} } which agree on ${\displaystyle \mathbb {Z} }$ are fact equal.

A proof sketch for example of non-surjective epimorphism in the category of rings by using zig-zag

We show that: Let ${\displaystyle \beta ,\gamma }$ to be ring homomorphisms, and ${\displaystyle n,m\in \mathbb {Z} }$, ${\displaystyle n\neq 0}$. When ${\displaystyle \beta (m)=\gamma (m)}$ for all ${\displaystyle m\in \mathbb {Z} }$, then ${\displaystyle \beta \left({\frac {m}{n}}\right)=\gamma \left({\frac {m}{n}}\right)}$ for all ${\displaystyle {\frac {m}{n}}\in \mathbb {Q} }$. ${\displaystyle {\begin{aligned}\beta \left({\frac {m}{n}}\right)&=\beta \left({\frac {1}{n}}\cdot m\right)=\beta \left({\frac {1}{n}}\right)\cdot \beta (m)\\&=\beta \left({\frac {1}{n}}\right)\cdot \gamma (m)=\beta \left({\frac {1}{n}}\right)\cdot \gamma \left(mn\cdot {\frac {1}{n}}\right)\\&=\beta \left({\frac {1}{n}}\right)\cdot \gamma (mn)\cdot \gamma \left({\frac {1}{n}}\right)=\beta \left({\frac {1}{n}}\right)\cdot \beta (mn)\cdot \gamma \left({\frac {1}{n}}\right)\\&=\beta \left({\frac {1}{n}}\cdot mn\right)\cdot \gamma \left({\frac {1}{n}}\right)=\beta (m)\cdot \gamma \left({\frac {1}{n}}\right)=\gamma (m)\cdot \gamma \left({\frac {1}{n}}\right)\\&=\gamma \left(m\cdot {\frac {1}{n}}\right)=\gamma \left({\frac {m}{n}}\right),\end{aligned}}}$ as required.

See also

• Epimorphisms