Mixing
Homomorphisms and semilattice decomposition of a band with an identity.
$begingroup$
My question is fairly simple. Suppose we have a right-zero semigroup with an attached identity and decompose it into a semilattice. How do we define the structural homomorphism from 1 into the right-zero semigroup? For instance, let $a,b$ be elements of the right zero semigroup suppose we map $1$ to $a$ then $b1=ba=a$ but $1$ is an identity.
Thanks.
edit: By clarity:
Let Y be the semilattice ${S_{alpha}: alpha in Y}$ of disjoint rectangular bands. For each pair $alpha, beta in Y$ $(alpha geq beta$) let $phi_{alpha, beta}:S_{alpha} longrightarrow S_{beta}$ be a homomorphism and suppose $phi_{alpha, alpha}$ is an automorphism of $S_{alpha}$ and $phi_{alpha, beta} phi_{beta, gamma} = phi_{alpha, gamma}$. Then the operation * on S is defined by $a_{alpha} * b_{beta} = (a_{alpha} phi_{alpha, alphabeta}) * (b_{beta} phi_{beta, alphabeta})$ for $a_{alpha} in S_{alpha}, b_{beta} in S_{beta}$.
Suppose $S_1$ is the rectangular band whose only element is 1. Then $1 geq alpha$ for all $alpha in Y$. If $S_{beta}$ is a, for example, 2 element right-zero semigroup, how do I define $phi_{1,beta}$?
Edit: The answer is you can't. I misunderstood the definition and these homomorphisms exist only in strong-semilattices. S, in this case, is not a strong semilattice.
semigroups
asked Feb 1 at 23:14
GlassGlass
112
$endgroup$
|
show 5 more comments
$begingroup$
My question is fairly simple. Suppose we have a right-zero semigroup with an attached identity and decompose it into a semilattice. How do we define the structural homomorphism from 1 into the right-zero semigroup? For instance, let $a,b$ be elements of the right zero semigroup suppose we map $1$ to $a$ then $b1=ba=a$ but $1$ is an identity.
Thanks.
edit: By clarity:
Let Y be the semilattice ${S_{alpha}: alpha in Y}$ of disjoint rectangular bands. For each pair $alpha, beta in Y$ $(alpha geq beta$) let $phi_{alpha, beta}:S_{alpha} longrightarrow S_{beta}$ be a homomorphism and suppose $phi_{alpha, alpha}$ is an automorphism of $S_{alpha}$ and $phi_{alpha, beta} phi_{beta, gamma} = phi_{alpha, gamma}$. Then the operation * on S is defined by $a_{alpha} * b_{beta} = (a_{alpha} phi_{alpha, alphabeta}) * (b_{beta} phi_{beta, alphabeta})$ for $a_{alpha} in S_{alpha}, b_{beta} in S_{beta}$.
Suppose $S_1$ is the rectangular band whose only element is 1. Then $1 geq alpha$ for all $alpha in Y$. If $S_{beta}$ is a, for example, 2 element right-zero semigroup, how do I define $phi_{1,beta}$?
Edit: The answer is you can't. I misunderstood the definition and these homomorphisms exist only in strong-semilattices. S, in this case, is not a strong semilattice.
semigroups
asked Feb 1 at 23:14
GlassGlass
112
$endgroup$
$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
1
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
1
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31
|
show 5 more comments
$begingroup$
My question is fairly simple. Suppose we have a right-zero semigroup with an attached identity and decompose it into a semilattice. How do we define the structural homomorphism from 1 into the right-zero semigroup? For instance, let $a,b$ be elements of the right zero semigroup suppose we map $1$ to $a$ then $b1=ba=a$ but $1$ is an identity.
Thanks.
edit: By clarity:
Let Y be the semilattice ${S_{alpha}: alpha in Y}$ of disjoint rectangular bands. For each pair $alpha, beta in Y$ $(alpha geq beta$) let $phi_{alpha, beta}:S_{alpha} longrightarrow S_{beta}$ be a homomorphism and suppose $phi_{alpha, alpha}$ is an automorphism of $S_{alpha}$ and $phi_{alpha, beta} phi_{beta, gamma} = phi_{alpha, gamma}$. Then the operation * on S is defined by $a_{alpha} * b_{beta} = (a_{alpha} phi_{alpha, alphabeta}) * (b_{beta} phi_{beta, alphabeta})$ for $a_{alpha} in S_{alpha}, b_{beta} in S_{beta}$.
Suppose $S_1$ is the rectangular band whose only element is 1. Then $1 geq alpha$ for all $alpha in Y$. If $S_{beta}$ is a, for example, 2 element right-zero semigroup, how do I define $phi_{1,beta}$?
Edit: The answer is you can't. I misunderstood the definition and these homomorphisms exist only in strong-semilattices. S, in this case, is not a strong semilattice.
semigroups
asked Feb 1 at 23:14
GlassGlass
112
$endgroup$
My question is fairly simple. Suppose we have a right-zero semigroup with an attached identity and decompose it into a semilattice. How do we define the structural homomorphism from 1 into the right-zero semigroup? For instance, let $a,b$ be elements of the right zero semigroup suppose we map $1$ to $a$ then $b1=ba=a$ but $1$ is an identity.
Thanks.
edit: By clarity:
Let Y be the semilattice ${S_{alpha}: alpha in Y}$ of disjoint rectangular bands. For each pair $alpha, beta in Y$ $(alpha geq beta$) let $phi_{alpha, beta}:S_{alpha} longrightarrow S_{beta}$ be a homomorphism and suppose $phi_{alpha, alpha}$ is an automorphism of $S_{alpha}$ and $phi_{alpha, beta} phi_{beta, gamma} = phi_{alpha, gamma}$. Then the operation * on S is defined by $a_{alpha} * b_{beta} = (a_{alpha} phi_{alpha, alphabeta}) * (b_{beta} phi_{beta, alphabeta})$ for $a_{alpha} in S_{alpha}, b_{beta} in S_{beta}$.
Suppose $S_1$ is the rectangular band whose only element is 1. Then $1 geq alpha$ for all $alpha in Y$. If $S_{beta}$ is a, for example, 2 element right-zero semigroup, how do I define $phi_{1,beta}$?
Edit: The answer is you can't. I misunderstood the definition and these homomorphisms exist only in strong-semilattices. S, in this case, is not a strong semilattice.
semigroups
semigroups
asked Feb 1 at 23:14
GlassGlass
112
asked Feb 1 at 23:14
GlassGlass
112
edited Feb 4 at 2:29
Glass
asked Feb 1 at 23:14
GlassGlass
112
asked Feb 1 at 23:14
GlassGlass
112
asked Feb 1 at 23:14
GlassGlass
112
112
$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
1
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
1
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31
|
show 5 more comments
$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
1
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
1
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31
$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
1
1
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
1
1
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31
|
show 5 more comments
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$begingroup$
What do you mean by 'structural homomorphism from 1'?
$endgroup$
– Berci
Feb 1 at 23:51
1
$begingroup$
I edited my original post for clarity.
$endgroup$
– Glass
Feb 2 at 0:33
$begingroup$
I think any map will work. We don't have to define these $phi_{a,b}$, but they are assumed to be given in order to draw the conclusion: there's a (certain kind of) semigroup structure arising on the disjoint union.
$endgroup$
– Berci
Feb 2 at 0:42
$begingroup$
But suppose $1 phi_{1,beta} = a$ then $b * 1 = b * 1 phi_{1,beta} = b*a=a$, but obviously $b*1=b$. What am I misunderstanding?
$endgroup$
– Glass
Feb 2 at 1:07
1
$begingroup$
$alpha$ is just part of the index set. Any band can be decomposed into a semilattice or rectangular bands which are in fact it's $mathscr{D}$-classes. If we attach an identity to a right zero semigroup we have two $mathscr{D}$- classes namely ${a,b}$ and ${1}$ which I have called $S_{beta}$ and $S_{1}$ respectively.
$endgroup$
– Glass
Feb 2 at 23:31