Semigroups with no morphisms between them Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Not a functor not prefunctorMorphisms with arbitrary number of arguments?Internalising the functor action on morphisms (e.g. to exponential objects)In a groupoid, do any two objects have a morphism between them?A reflective subcategory of the category of inverse semigroups.Is every semigroup with (possibly non-unique) division a group?Examples of left-topological compact semigroupsProblems with subcategoriesInclusion relations between equationally defined classes of finite semigroupsDo empty category objects have self morphisms?
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Semigroups with no morphisms between them
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Semigroups with no morphisms between them
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Not a functor not prefunctorMorphisms with arbitrary number of arguments?Internalising the functor action on morphisms (e.g. to exponential objects)In a groupoid, do any two objects have a morphism between them?A reflective subcategory of the category of inverse semigroups.Is every semigroup with (possibly non-unique) division a group?Examples of left-topological compact semigroupsProblems with subcategoriesInclusion relations between equationally defined classes of finite semigroupsDo empty category objects have self morphisms?
$begingroup$
Given two monoids we always have a morphism from one to the other thanks to the presence of the identity element.
Are there examples of non-empty semigroups that have no morphisms from one to the other? The destination semigroup can't be finite, because if we have an idempotent present, we just map everything to it and get a constant morphism. Apparently, it can't contain an idempotent, period.
Put another way, the subcategory of finite semigroups is clearly strongly connected. Is the same true of the category of all semigroups?
Are there two such non-empty semigroups that don't have a morphism in either direction?
category-theory examples-counterexamples semigroups
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
$endgroup$
add a comment |
$begingroup$
Given two monoids we always have a morphism from one to the other thanks to the presence of the identity element.
Are there examples of non-empty semigroups that have no morphisms from one to the other? The destination semigroup can't be finite, because if we have an idempotent present, we just map everything to it and get a constant morphism. Apparently, it can't contain an idempotent, period.
Put another way, the subcategory of finite semigroups is clearly strongly connected. Is the same true of the category of all semigroups?
Are there two such non-empty semigroups that don't have a morphism in either direction?
category-theory examples-counterexamples semigroups
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
$endgroup$
$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago
add a comment |
$begingroup$
Given two monoids we always have a morphism from one to the other thanks to the presence of the identity element.
Are there examples of non-empty semigroups that have no morphisms from one to the other? The destination semigroup can't be finite, because if we have an idempotent present, we just map everything to it and get a constant morphism. Apparently, it can't contain an idempotent, period.
Put another way, the subcategory of finite semigroups is clearly strongly connected. Is the same true of the category of all semigroups?
Are there two such non-empty semigroups that don't have a morphism in either direction?
category-theory examples-counterexamples semigroups
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
$endgroup$
Given two monoids we always have a morphism from one to the other thanks to the presence of the identity element.
Are there examples of non-empty semigroups that have no morphisms from one to the other? The destination semigroup can't be finite, because if we have an idempotent present, we just map everything to it and get a constant morphism. Apparently, it can't contain an idempotent, period.
Put another way, the subcategory of finite semigroups is clearly strongly connected. Is the same true of the category of all semigroups?
Are there two such non-empty semigroups that don't have a morphism in either direction?
category-theory examples-counterexamples semigroups
category-theory examples-counterexamples semigroups
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
edited 1 hour ago
Alvin Lepik
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
asked 1 hour ago
Alvin LepikAlvin Lepik
2,92511024
2,92511024
$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago
add a comment |
$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A to mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n in mathbb N^+$. Now trying to add an element $a in A$ to itself $n+1$ times cannot be respected by the map:
$$
f(underbracea + ldots + a_n+1 text times) = underbracef(a) + ldots + f(a)_n+1 text times geq n+1,
$$
which contradicts that the range of $f$ is bounded by $n$.
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
$endgroup$
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
add a comment |
$begingroup$
There's no morphism from $0$ (or from any semigroup with idempotent) to the additive semigroup on $2,4,6,dots$
answered 1 hour ago
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A to mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n in mathbb N^+$. Now trying to add an element $a in A$ to itself $n+1$ times cannot be respected by the map:
$$
f(underbracea + ldots + a_n+1 text times) = underbracef(a) + ldots + f(a)_n+1 text times geq n+1,
$$
which contradicts that the range of $f$ is bounded by $n$.
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
$endgroup$
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
add a comment |
$begingroup$
Let $mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A to mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n in mathbb N^+$. Now trying to add an element $a in A$ to itself $n+1$ times cannot be respected by the map:
$$
f(underbracea + ldots + a_n+1 text times) = underbracef(a) + ldots + f(a)_n+1 text times geq n+1,
$$
which contradicts that the range of $f$ is bounded by $n$.
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
$endgroup$
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
add a comment |
$begingroup$
Let $mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A to mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n in mathbb N^+$. Now trying to add an element $a in A$ to itself $n+1$ times cannot be respected by the map:
$$
f(underbracea + ldots + a_n+1 text times) = underbracef(a) + ldots + f(a)_n+1 text times geq n+1,
$$
which contradicts that the range of $f$ is bounded by $n$.
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
$endgroup$
Let $mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A to mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n in mathbb N^+$. Now trying to add an element $a in A$ to itself $n+1$ times cannot be respected by the map:
$$
f(underbracea + ldots + a_n+1 text times) = underbracef(a) + ldots + f(a)_n+1 text times geq n+1,
$$
which contradicts that the range of $f$ is bounded by $n$.
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
answered 1 hour ago
Mark KamsmaMark Kamsma
79311
79311
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
add a comment |
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
$begingroup$
But as stated in the question, $A$ being finite means there is a morphism in the other direction
$endgroup$
– Max
12 mins ago
add a comment |
$begingroup$
There's no morphism from $0$ (or from any semigroup with idempotent) to the additive semigroup on $2,4,6,dots$
answered 1 hour ago
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
add a comment |
$begingroup$
There's no morphism from $0$ (or from any semigroup with idempotent) to the additive semigroup on $2,4,6,dots$
answered 1 hour ago
BerciBerci
62.1k23776
$endgroup$
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
add a comment |
$begingroup$
There's no morphism from $0$ (or from any semigroup with idempotent) to the additive semigroup on $2,4,6,dots$
answered 1 hour ago
BerciBerci
62.1k23776
$endgroup$
There's no morphism from $0$ (or from any semigroup with idempotent) to the additive semigroup on $2,4,6,dots$
answered 1 hour ago
BerciBerci
62.1k23776
answered 1 hour ago
BerciBerci
62.1k23776
answered 1 hour ago
BerciBerci
62.1k23776
answered 1 hour ago
BerciBerci
62.1k23776
62.1k23776
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
add a comment |
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
Why did you pick $2,4,6,dots$ instead of $1,2,3,dots$? (Just curious)
$endgroup$
– Alex Kruckman
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
@AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to
$endgroup$
– Alvin Lepik
1 hour ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
$begingroup$
Yes. Well, my first thought was that the ring $2Bbb Z$ has unusual ring properties..
$endgroup$
– Berci
29 mins ago
add a comment |
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$begingroup$
@DanielSchepler ah yes, good catch, I am interested in the non-empty case.
$endgroup$
– Alvin Lepik
1 hour ago