How many endomorphisms $(mathbb{N},times)$ has? Like object in Mon category
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I was told that there is only one morphism in $Mon$ category for this object $(mathbb{N},times)$. But why?
I think that we can write every natural number as the product of a certain set of prime numbers. So we can say that the set of primes forms the basis of this monoid $(mathbb{N},times)$, that is, linear combinations with positive integer coefficients (powers) make up the entire set of supports for the structure of this monoid: for any $ninmathbb{N}$, $n = p_1^{g_1}timescdotstimes p_i^{g_i}$, where $forall p_j in {p_1,p_2,cdots}$ -- basis, and $forall g_jin mathbb{Z_{>0}}$. And we can make some transpositions on the elements of the basis ${p_1,p_2,cdots}$ thereby defining the morphism (where always $1$ going to $1$ by this morphism). So we have infinite number of morphisms, I am right?
abstract-algebra category-theory monoid
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
$endgroup$
add a comment |
$begingroup$
I was told that there is only one morphism in $Mon$ category for this object $(mathbb{N},times)$. But why?
I think that we can write every natural number as the product of a certain set of prime numbers. So we can say that the set of primes forms the basis of this monoid $(mathbb{N},times)$, that is, linear combinations with positive integer coefficients (powers) make up the entire set of supports for the structure of this monoid: for any $ninmathbb{N}$, $n = p_1^{g_1}timescdotstimes p_i^{g_i}$, where $forall p_j in {p_1,p_2,cdots}$ -- basis, and $forall g_jin mathbb{Z_{>0}}$. And we can make some transpositions on the elements of the basis ${p_1,p_2,cdots}$ thereby defining the morphism (where always $1$ going to $1$ by this morphism). So we have infinite number of morphisms, I am right?
abstract-algebra category-theory monoid
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
$endgroup$
3
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
2
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43
add a comment |
$begingroup$
I was told that there is only one morphism in $Mon$ category for this object $(mathbb{N},times)$. But why?
I think that we can write every natural number as the product of a certain set of prime numbers. So we can say that the set of primes forms the basis of this monoid $(mathbb{N},times)$, that is, linear combinations with positive integer coefficients (powers) make up the entire set of supports for the structure of this monoid: for any $ninmathbb{N}$, $n = p_1^{g_1}timescdotstimes p_i^{g_i}$, where $forall p_j in {p_1,p_2,cdots}$ -- basis, and $forall g_jin mathbb{Z_{>0}}$. And we can make some transpositions on the elements of the basis ${p_1,p_2,cdots}$ thereby defining the morphism (where always $1$ going to $1$ by this morphism). So we have infinite number of morphisms, I am right?
abstract-algebra category-theory monoid
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
$endgroup$
I was told that there is only one morphism in $Mon$ category for this object $(mathbb{N},times)$. But why?
I think that we can write every natural number as the product of a certain set of prime numbers. So we can say that the set of primes forms the basis of this monoid $(mathbb{N},times)$, that is, linear combinations with positive integer coefficients (powers) make up the entire set of supports for the structure of this monoid: for any $ninmathbb{N}$, $n = p_1^{g_1}timescdotstimes p_i^{g_i}$, where $forall p_j in {p_1,p_2,cdots}$ -- basis, and $forall g_jin mathbb{Z_{>0}}$. And we can make some transpositions on the elements of the basis ${p_1,p_2,cdots}$ thereby defining the morphism (where always $1$ going to $1$ by this morphism). So we have infinite number of morphisms, I am right?
abstract-algebra category-theory monoid
abstract-algebra category-theory monoid
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
edited Dec 16 '18 at 4:24
Just do it
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
asked Dec 16 '18 at 2:22
Just do itJust do it
19618
19618
3
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
2
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43
add a comment |
3
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
2
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43
3
3
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
2
2
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43
add a comment |
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3
$begingroup$
Are you sure they said $(mathbb{N},times)$ and not $(mathbb{N},+)$?
$endgroup$
– Eric Wofsey
Dec 16 '18 at 2:23
$begingroup$
@EricWofsey ye, multiplication
$endgroup$
– Just do it
Dec 16 '18 at 2:32
2
$begingroup$
Yes an infinite number of morphisms
$endgroup$
– Hermit with Adjoint
Dec 16 '18 at 3:43