Why are epimorphisms in the category of Monoids not necessarily surjections?
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The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?
category-theory
asked Feb 1 at 17:40
user56834user56834
3,10521252
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add a comment |
$begingroup$
The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?
category-theory
asked Feb 1 at 17:40
user56834user56834
3,10521252
$endgroup$
add a comment |
$begingroup$
The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?
category-theory
asked Feb 1 at 17:40
user56834user56834
3,10521252
$endgroup$
The book "basic category theory" states that in the category Mon, epimorphisms are not necessarily surjections, but doesn't explain why. Why is this the case?
category-theory
category-theory
asked Feb 1 at 17:40
user56834user56834
3,10521252
asked Feb 1 at 17:40
user56834user56834
3,10521252
asked Feb 1 at 17:40
user56834user56834
3,10521252
asked Feb 1 at 17:40
user56834user56834
3,10521252
asked Feb 1 at 17:40
user56834user56834
3,10521252
3,10521252
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
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add a comment |
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Quoting Wikipedia, we see:
In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.
answered Feb 1 at 17:46
vadim123vadim123
76k897189
$endgroup$
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
add a comment |
$begingroup$
Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.
answered Feb 1 at 17:47
user3482749user3482749
4,266919
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
Let $Bbb Z$ be the additive group of integers, and $Bbb N_0={0,1,2,ldots}$.
Then the inclusion $Bbb N_0toBbb Z$ is an epimorphism in Mon.
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Feb 1 at 17:45
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
$begingroup$
Quoting Wikipedia, we see:
In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.
answered Feb 1 at 17:46
vadim123vadim123
76k897189
$endgroup$
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
add a comment |
$begingroup$
Quoting Wikipedia, we see:
In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.
answered Feb 1 at 17:46
vadim123vadim123
76k897189
$endgroup$
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
add a comment |
$begingroup$
Quoting Wikipedia, we see:
In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.
answered Feb 1 at 17:46
vadim123vadim123
76k897189
$endgroup$
Quoting Wikipedia, we see:
In the category of monoids, Mon, the inclusion map $mathbb{N}tomathbb{Z}$ is a non-surjective epimorphism. To see this, suppose that $g_1$ and $g_2$ are two distinct maps from $mathbb{Z}$ to some monoid $M$. Then for some $ninmathbb{Z}$, $g_1(n)neq g_2(n)$, so $g_1(-n)neq g_2(-n)$. Either $n$ or $-n$ is in $mathbb{N}$,, so the restrictions of $g_1$ and $g_2$ to $mathbb{N}$ are unequal.
answered Feb 1 at 17:46
vadim123vadim123
76k897189
answered Feb 1 at 17:46
vadim123vadim123
76k897189
answered Feb 1 at 17:46
vadim123vadim123
76k897189
answered Feb 1 at 17:46
vadim123vadim123
76k897189
76k897189
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
add a comment |
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
$begingroup$
I don't understand the argument? specifically, why does $g_1(-n)neq g_2(-n)$ need to hold? That would hold if we were in the category of groups, I see that. But I don't see why it would hold in the category of monoids. also, I don't get the last sentence.
$endgroup$
– user56834
Feb 1 at 18:17
1
1
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
$begingroup$
$0=g_1(0)=g_1(n+(-n))=g_1(n)+g_1(-n)$
$endgroup$
– vadim123
Feb 1 at 20:53
add a comment |
$begingroup$
Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.
answered Feb 1 at 17:47
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.
answered Feb 1 at 17:47
user3482749user3482749
4,266919
$endgroup$
add a comment |
$begingroup$
Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.
answered Feb 1 at 17:47
user3482749user3482749
4,266919
$endgroup$
Consider the inclusion $i: mathbb{N}hookrightarrowmathbb{Z}$. This is clearly not surjective. However, for any monoid $M$ and any two maps $f,g:mathbb{Z} to M$ such that $f circ i = g circ i$, we must have $f = g$, by a simple argument.
answered Feb 1 at 17:47
user3482749user3482749
4,266919
answered Feb 1 at 17:47
user3482749user3482749
4,266919
answered Feb 1 at 17:47
user3482749user3482749
4,266919
answered Feb 1 at 17:47
user3482749user3482749
4,266919
4,266919
add a comment |
add a comment |
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