Counterexample for Converse of Surjective Homomorphisms
$begingroup$
In universal algebra, I am trying to find a counterexample using groups for the converse of the following:
If $mathcal{A},mathcal{B}$ are algebras, and $phi:mathcal{A}to mathcal{B}$ is a surjective homomorphism and the identity $mathbf{s}=mathbf{t}$ holds in $mathcal{A}$, then it also holds in $mathcal{B}$.
I was thinking the trivial homomorphism could possibly be a counterexample: $G$ any non-abelian group, $H$ any abelian group, $phi: G to H$ defined by $phi(g)=1$ and the identity can be commutativity. If $H$ is not trivial, then $phi$ is not surjective.
Is my counterexample too complicated or wrong? Any advice on how to find a simple one?
group-homomorphism universal-algebra
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
$endgroup$
|
show 4 more comments
$begingroup$
In universal algebra, I am trying to find a counterexample using groups for the converse of the following:
If $mathcal{A},mathcal{B}$ are algebras, and $phi:mathcal{A}to mathcal{B}$ is a surjective homomorphism and the identity $mathbf{s}=mathbf{t}$ holds in $mathcal{A}$, then it also holds in $mathcal{B}$.
I was thinking the trivial homomorphism could possibly be a counterexample: $G$ any non-abelian group, $H$ any abelian group, $phi: G to H$ defined by $phi(g)=1$ and the identity can be commutativity. If $H$ is not trivial, then $phi$ is not surjective.
Is my counterexample too complicated or wrong? Any advice on how to find a simple one?
group-homomorphism universal-algebra
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
$endgroup$
$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
1
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
1
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
2
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
1
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35
|
show 4 more comments
$begingroup$
In universal algebra, I am trying to find a counterexample using groups for the converse of the following:
If $mathcal{A},mathcal{B}$ are algebras, and $phi:mathcal{A}to mathcal{B}$ is a surjective homomorphism and the identity $mathbf{s}=mathbf{t}$ holds in $mathcal{A}$, then it also holds in $mathcal{B}$.
I was thinking the trivial homomorphism could possibly be a counterexample: $G$ any non-abelian group, $H$ any abelian group, $phi: G to H$ defined by $phi(g)=1$ and the identity can be commutativity. If $H$ is not trivial, then $phi$ is not surjective.
Is my counterexample too complicated or wrong? Any advice on how to find a simple one?
group-homomorphism universal-algebra
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
$endgroup$
In universal algebra, I am trying to find a counterexample using groups for the converse of the following:
If $mathcal{A},mathcal{B}$ are algebras, and $phi:mathcal{A}to mathcal{B}$ is a surjective homomorphism and the identity $mathbf{s}=mathbf{t}$ holds in $mathcal{A}$, then it also holds in $mathcal{B}$.
I was thinking the trivial homomorphism could possibly be a counterexample: $G$ any non-abelian group, $H$ any abelian group, $phi: G to H$ defined by $phi(g)=1$ and the identity can be commutativity. If $H$ is not trivial, then $phi$ is not surjective.
Is my counterexample too complicated or wrong? Any advice on how to find a simple one?
group-homomorphism universal-algebra
group-homomorphism universal-algebra
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
asked Jan 18 at 3:56
numericalorangenumericalorange
1,755311
1,755311
$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
1
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
1
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
2
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
1
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35
|
show 4 more comments
$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
1
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
1
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
2
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
1
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35
$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
1
1
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
1
1
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
2
2
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
1
1
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35
|
show 4 more comments
1 Answer
1
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oldest
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$begingroup$
For any kind of algebra (regardless of the operations it has) the converse of that result is indeed false.
Just consider that if $mathbf A$ is an algebra, then $theta = A^2$ is a congruence on $mathbf A$, and the quotient $mathbf A/theta$ is a one-element algebra.
Now, one-element algebras satisfy the equation $x=y$, and they're the only ones that do.
So if $mathbf A$ is a non-trivial algebra (if it has more than one element), you can consider $mathbf B = mathbf A/theta$, and the only possible map $phi:mathbf A to mathbf B$ is a surjective homomorphism, showing that the converse of the result is false.
Notice that it is necessary that $mathbf A$ is non-trivial, since trivial algebras (trivially) satisfy any equation whatever.
Now you can just as well replace $x=y$ by any equation that you know that the non-trivial algebra $mathbf A$ doesn't satisfy (because $mathbf A/theta$ always does).
In particular, if $mathbf A$ is a non-commutative group, you can consider the identity $xy=yx$, as you did.
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
$endgroup$
add a comment |
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$begingroup$
For any kind of algebra (regardless of the operations it has) the converse of that result is indeed false.
Just consider that if $mathbf A$ is an algebra, then $theta = A^2$ is a congruence on $mathbf A$, and the quotient $mathbf A/theta$ is a one-element algebra.
Now, one-element algebras satisfy the equation $x=y$, and they're the only ones that do.
So if $mathbf A$ is a non-trivial algebra (if it has more than one element), you can consider $mathbf B = mathbf A/theta$, and the only possible map $phi:mathbf A to mathbf B$ is a surjective homomorphism, showing that the converse of the result is false.
Notice that it is necessary that $mathbf A$ is non-trivial, since trivial algebras (trivially) satisfy any equation whatever.
Now you can just as well replace $x=y$ by any equation that you know that the non-trivial algebra $mathbf A$ doesn't satisfy (because $mathbf A/theta$ always does).
In particular, if $mathbf A$ is a non-commutative group, you can consider the identity $xy=yx$, as you did.
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
$endgroup$
add a comment |
$begingroup$
For any kind of algebra (regardless of the operations it has) the converse of that result is indeed false.
Just consider that if $mathbf A$ is an algebra, then $theta = A^2$ is a congruence on $mathbf A$, and the quotient $mathbf A/theta$ is a one-element algebra.
Now, one-element algebras satisfy the equation $x=y$, and they're the only ones that do.
So if $mathbf A$ is a non-trivial algebra (if it has more than one element), you can consider $mathbf B = mathbf A/theta$, and the only possible map $phi:mathbf A to mathbf B$ is a surjective homomorphism, showing that the converse of the result is false.
Notice that it is necessary that $mathbf A$ is non-trivial, since trivial algebras (trivially) satisfy any equation whatever.
Now you can just as well replace $x=y$ by any equation that you know that the non-trivial algebra $mathbf A$ doesn't satisfy (because $mathbf A/theta$ always does).
In particular, if $mathbf A$ is a non-commutative group, you can consider the identity $xy=yx$, as you did.
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
$endgroup$
add a comment |
$begingroup$
For any kind of algebra (regardless of the operations it has) the converse of that result is indeed false.
Just consider that if $mathbf A$ is an algebra, then $theta = A^2$ is a congruence on $mathbf A$, and the quotient $mathbf A/theta$ is a one-element algebra.
Now, one-element algebras satisfy the equation $x=y$, and they're the only ones that do.
So if $mathbf A$ is a non-trivial algebra (if it has more than one element), you can consider $mathbf B = mathbf A/theta$, and the only possible map $phi:mathbf A to mathbf B$ is a surjective homomorphism, showing that the converse of the result is false.
Notice that it is necessary that $mathbf A$ is non-trivial, since trivial algebras (trivially) satisfy any equation whatever.
Now you can just as well replace $x=y$ by any equation that you know that the non-trivial algebra $mathbf A$ doesn't satisfy (because $mathbf A/theta$ always does).
In particular, if $mathbf A$ is a non-commutative group, you can consider the identity $xy=yx$, as you did.
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
$endgroup$
For any kind of algebra (regardless of the operations it has) the converse of that result is indeed false.
Just consider that if $mathbf A$ is an algebra, then $theta = A^2$ is a congruence on $mathbf A$, and the quotient $mathbf A/theta$ is a one-element algebra.
Now, one-element algebras satisfy the equation $x=y$, and they're the only ones that do.
So if $mathbf A$ is a non-trivial algebra (if it has more than one element), you can consider $mathbf B = mathbf A/theta$, and the only possible map $phi:mathbf A to mathbf B$ is a surjective homomorphism, showing that the converse of the result is false.
Notice that it is necessary that $mathbf A$ is non-trivial, since trivial algebras (trivially) satisfy any equation whatever.
Now you can just as well replace $x=y$ by any equation that you know that the non-trivial algebra $mathbf A$ doesn't satisfy (because $mathbf A/theta$ always does).
In particular, if $mathbf A$ is a non-commutative group, you can consider the identity $xy=yx$, as you did.
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
answered Jan 20 at 18:23
amrsaamrsa
3,6052618
3,6052618
add a comment |
add a comment |
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$begingroup$
But the statement is correct; it's very easy to see that it is
$endgroup$
– Max
Jan 18 at 9:11
1
$begingroup$
Your counterexample is right. The identity $xy=yx$ holds trivially in the trivial group, but not in any non-abelian group. The comment that follows (if $H$ is not trivial...) seems redundant, since, by hypothesis, $phi$ is surjective; it just tells us that with that trivial homomorphism you only have a counterexample to the converse of the result if the group it maps to is also trivial.
$endgroup$
– amrsa
Jan 18 at 9:52
1
$begingroup$
@Max I think you misread the question. The goal is to find a counterexample to the converse of the statement. Thus, an identity that holds in $mathcal B$ but not in $mathcal A$, while $phi:mathcal A tomathcal B$ is surjective.
$endgroup$
– amrsa
Jan 18 at 9:55
2
$begingroup$
It should also be pointed that you don't need the specific case of groups, since a counterexample can be found for any non-trivial $mathcal A$: just consider the natural homomorphism $nu_{theta}:mathcal Atomathcal A/theta$, where $theta = A^2$ and the identity is $x=y$.
$endgroup$
– amrsa
Jan 18 at 9:57
1
$begingroup$
@PedroSánchezTerraf Thanks for the encouragement :) The comment wasn't a straight answer to the question. I made it into an answer and noted how it can be changed to make the OP solution a particular case (of the changed version).
$endgroup$
– amrsa
Jan 20 at 18:35