Why does $mathbbZ$ represent the forgetful functor $U:mathbfGrptomathbfSet$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)$textGL_n:mathbfAlg_krightarrowmathbfSet:Rmapsto textGL_n(R)$ is a representable functorDualizing module and finiteness hypothesisUnderlying set of the free monoid, does it contain the empty string?checking the functor $textttNil_n$ is represented by $(mathbbZ[x]/(x^n), tau_R)$Composition of monadic functors isn't monadicShowing the Affine functor $underlinemathbbA^r$ is representable by Affine Space $mathbbA^r := Spec(mathbbZ[X_1, dots, X_r])$Show that the functor that takes $R$ to the set of invertible elements of $R[X]/(X^2-a)$ is representable.
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Why does $mathbbZ$ represent the forgetful functor $U:mathbfGrptomathbfSet$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)$textGL_n:mathbfAlg_krightarrowmathbfSet:Rmapsto textGL_n(R)$ is a representable functorDualizing module and finiteness hypothesisUnderlying set of the free monoid, does it contain the empty string?checking the functor $textttNil_n$ is represented by $(mathbbZ[x]/(x^n), tau_R)$Composition of monadic functors isn't monadicShowing the Affine functor $underlinemathbbA^r$ is representable by Affine Space $mathbbA^r := Spec(mathbbZ[X_1, dots, X_r])$Show that the functor that takes $R$ to the set of invertible elements of $R[X]/(X^2-a)$ is representable.
$begingroup$
This is from Emily Riehl's Category theory in context
The forgetful functor $U:mathbfGrptomathbfSet$ is represented by the group $mathbbZ$ thanks to the natural isomorphism $alpha:mathbfGrp(mathbbZ,-)cong U$ whose components are the isomorphisms
$$beginalign
alpha_G:mathbfGrp(mathbbZ,G) & longrightarrow UG\
big[f:mathbbZto Gbig] & longmapsto f(1)
endalign$$
My impression is that if some group $A$ were to represent the functor $U$ it should have some distinguished element we could attach all the information to. If the group $A$ doesn't have such element then the only "reasonable" map $mathbfGrp(A,G)to UG$ would be $big[f:Ato Gbig]mapsto f(id_A)$ and that does not define an isomorphism. The only groups having a "distinguished element" that come to mind are the cyclic groups and their generators.
But then, why $mathbbZ$? Doesn't every (non trivial) cyclic group work? Given the cyclic group $mathbbZ_n$, the components
$$beginalign
alpha_A:mathbfGrp(mathbbZ_n,G) & longrightarrow UG\
big[f:mathbbZ_nto Gbig] & longmapsto f(1)
endalign$$
are isomorphisms: if $f(1)=alpha_A(f)=alpha_A(g)=g(1)$ then, for every $minmathbbZ_n$ $f(m)=f(1)+cdots+f(1)=g(1)+cdots+g(1)=g(m)$ so that $gequiv f$ and for every $gin UG$ there exists a unique $f_g:mathbbZ_nto G$ such that $f_g(1)=g$.
Where am I wrong? Could it be that $mathbbZ=langle1rangle$ is the only cyclic group with only one generator while $mathbbZ_n=langle1rangle=langle n-1rangle$ for every $n$? If so, why? It doesn't seem very important to the problem.
Thanks in advance
representable-functor forgetful-functors
asked Mar 27 at 20:34
PedroPedro
541212
$endgroup$
add a comment |
$begingroup$
This is from Emily Riehl's Category theory in context
The forgetful functor $U:mathbfGrptomathbfSet$ is represented by the group $mathbbZ$ thanks to the natural isomorphism $alpha:mathbfGrp(mathbbZ,-)cong U$ whose components are the isomorphisms
$$beginalign
alpha_G:mathbfGrp(mathbbZ,G) & longrightarrow UG\
big[f:mathbbZto Gbig] & longmapsto f(1)
endalign$$
My impression is that if some group $A$ were to represent the functor $U$ it should have some distinguished element we could attach all the information to. If the group $A$ doesn't have such element then the only "reasonable" map $mathbfGrp(A,G)to UG$ would be $big[f:Ato Gbig]mapsto f(id_A)$ and that does not define an isomorphism. The only groups having a "distinguished element" that come to mind are the cyclic groups and their generators.
But then, why $mathbbZ$? Doesn't every (non trivial) cyclic group work? Given the cyclic group $mathbbZ_n$, the components
$$beginalign
alpha_A:mathbfGrp(mathbbZ_n,G) & longrightarrow UG\
big[f:mathbbZ_nto Gbig] & longmapsto f(1)
endalign$$
are isomorphisms: if $f(1)=alpha_A(f)=alpha_A(g)=g(1)$ then, for every $minmathbbZ_n$ $f(m)=f(1)+cdots+f(1)=g(1)+cdots+g(1)=g(m)$ so that $gequiv f$ and for every $gin UG$ there exists a unique $f_g:mathbbZ_nto G$ such that $f_g(1)=g$.
Where am I wrong? Could it be that $mathbbZ=langle1rangle$ is the only cyclic group with only one generator while $mathbbZ_n=langle1rangle=langle n-1rangle$ for every $n$? If so, why? It doesn't seem very important to the problem.
Thanks in advance
representable-functor forgetful-functors
asked Mar 27 at 20:34
PedroPedro
541212
$endgroup$
$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42
add a comment |
$begingroup$
This is from Emily Riehl's Category theory in context
The forgetful functor $U:mathbfGrptomathbfSet$ is represented by the group $mathbbZ$ thanks to the natural isomorphism $alpha:mathbfGrp(mathbbZ,-)cong U$ whose components are the isomorphisms
$$beginalign
alpha_G:mathbfGrp(mathbbZ,G) & longrightarrow UG\
big[f:mathbbZto Gbig] & longmapsto f(1)
endalign$$
My impression is that if some group $A$ were to represent the functor $U$ it should have some distinguished element we could attach all the information to. If the group $A$ doesn't have such element then the only "reasonable" map $mathbfGrp(A,G)to UG$ would be $big[f:Ato Gbig]mapsto f(id_A)$ and that does not define an isomorphism. The only groups having a "distinguished element" that come to mind are the cyclic groups and their generators.
But then, why $mathbbZ$? Doesn't every (non trivial) cyclic group work? Given the cyclic group $mathbbZ_n$, the components
$$beginalign
alpha_A:mathbfGrp(mathbbZ_n,G) & longrightarrow UG\
big[f:mathbbZ_nto Gbig] & longmapsto f(1)
endalign$$
are isomorphisms: if $f(1)=alpha_A(f)=alpha_A(g)=g(1)$ then, for every $minmathbbZ_n$ $f(m)=f(1)+cdots+f(1)=g(1)+cdots+g(1)=g(m)$ so that $gequiv f$ and for every $gin UG$ there exists a unique $f_g:mathbbZ_nto G$ such that $f_g(1)=g$.
Where am I wrong? Could it be that $mathbbZ=langle1rangle$ is the only cyclic group with only one generator while $mathbbZ_n=langle1rangle=langle n-1rangle$ for every $n$? If so, why? It doesn't seem very important to the problem.
Thanks in advance
representable-functor forgetful-functors
asked Mar 27 at 20:34
PedroPedro
541212
$endgroup$
This is from Emily Riehl's Category theory in context
The forgetful functor $U:mathbfGrptomathbfSet$ is represented by the group $mathbbZ$ thanks to the natural isomorphism $alpha:mathbfGrp(mathbbZ,-)cong U$ whose components are the isomorphisms
$$beginalign
alpha_G:mathbfGrp(mathbbZ,G) & longrightarrow UG\
big[f:mathbbZto Gbig] & longmapsto f(1)
endalign$$
My impression is that if some group $A$ were to represent the functor $U$ it should have some distinguished element we could attach all the information to. If the group $A$ doesn't have such element then the only "reasonable" map $mathbfGrp(A,G)to UG$ would be $big[f:Ato Gbig]mapsto f(id_A)$ and that does not define an isomorphism. The only groups having a "distinguished element" that come to mind are the cyclic groups and their generators.
But then, why $mathbbZ$? Doesn't every (non trivial) cyclic group work? Given the cyclic group $mathbbZ_n$, the components
$$beginalign
alpha_A:mathbfGrp(mathbbZ_n,G) & longrightarrow UG\
big[f:mathbbZ_nto Gbig] & longmapsto f(1)
endalign$$
are isomorphisms: if $f(1)=alpha_A(f)=alpha_A(g)=g(1)$ then, for every $minmathbbZ_n$ $f(m)=f(1)+cdots+f(1)=g(1)+cdots+g(1)=g(m)$ so that $gequiv f$ and for every $gin UG$ there exists a unique $f_g:mathbbZ_nto G$ such that $f_g(1)=g$.
Where am I wrong? Could it be that $mathbbZ=langle1rangle$ is the only cyclic group with only one generator while $mathbbZ_n=langle1rangle=langle n-1rangle$ for every $n$? If so, why? It doesn't seem very important to the problem.
Thanks in advance
representable-functor forgetful-functors
representable-functor forgetful-functors
asked Mar 27 at 20:34
PedroPedro
541212
asked Mar 27 at 20:34
PedroPedro
541212
asked Mar 27 at 20:34
PedroPedro
541212
asked Mar 27 at 20:34
PedroPedro
541212
asked Mar 27 at 20:34
PedroPedro
541212
541212
$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42
add a comment |
$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42
$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42
$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take $G=Bbb Z$. Then the only element of $bf Grp(BbbZ_m,Bbb Z)$
is the zero map. But $UBbb Z$ has rather more than one element.
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
add a comment |
$begingroup$
Let $S_1 = 1 $. Then you have a natural bijection between $mathbfSet(S_1,UG)$ and $UG$. Moreover, for any set $S$, you have a natural bijection between $mathbfSet(S,UG)$ and $mathbfGrp(F(S),G)$, where $F(S)$ is the free group with basis $S$. This is the universal property of the free group generated by $S$. But $F(S_1)$ is nothing else than the infinite cyclic group with one generator $1$.
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
active
oldest
votes
$begingroup$
Take $G=Bbb Z$. Then the only element of $bf Grp(BbbZ_m,Bbb Z)$
is the zero map. But $UBbb Z$ has rather more than one element.
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
add a comment |
$begingroup$
Take $G=Bbb Z$. Then the only element of $bf Grp(BbbZ_m,Bbb Z)$
is the zero map. But $UBbb Z$ has rather more than one element.
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
add a comment |
$begingroup$
Take $G=Bbb Z$. Then the only element of $bf Grp(BbbZ_m,Bbb Z)$
is the zero map. But $UBbb Z$ has rather more than one element.
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
Take $G=Bbb Z$. Then the only element of $bf Grp(BbbZ_m,Bbb Z)$
is the zero map. But $UBbb Z$ has rather more than one element.
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 20:41
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
add a comment |
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
$begingroup$
so the problem is that defining only the image of $1inmathbbZ_n$ does not necessarily define a homomorphism $mathbbZ_nto$something ??
$endgroup$
– Pedro
Mar 27 at 20:58
1
1
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
$begingroup$
@Pedro Exactly. For a homomorphism $f: mathbb Z_n rightarrow G$ the image of $1 in mathbb Z_n$ should have order dividing $n$, for $ncdot f(1) = f(ncdot 1) = f(0)$.
$endgroup$
– lisyarus
Mar 27 at 22:54
add a comment |
$begingroup$
Let $S_1 = 1 $. Then you have a natural bijection between $mathbfSet(S_1,UG)$ and $UG$. Moreover, for any set $S$, you have a natural bijection between $mathbfSet(S,UG)$ and $mathbfGrp(F(S),G)$, where $F(S)$ is the free group with basis $S$. This is the universal property of the free group generated by $S$. But $F(S_1)$ is nothing else than the infinite cyclic group with one generator $1$.
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
$begingroup$
Let $S_1 = 1 $. Then you have a natural bijection between $mathbfSet(S_1,UG)$ and $UG$. Moreover, for any set $S$, you have a natural bijection between $mathbfSet(S,UG)$ and $mathbfGrp(F(S),G)$, where $F(S)$ is the free group with basis $S$. This is the universal property of the free group generated by $S$. But $F(S_1)$ is nothing else than the infinite cyclic group with one generator $1$.
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
$endgroup$
add a comment |
$begingroup$
Let $S_1 = 1 $. Then you have a natural bijection between $mathbfSet(S_1,UG)$ and $UG$. Moreover, for any set $S$, you have a natural bijection between $mathbfSet(S,UG)$ and $mathbfGrp(F(S),G)$, where $F(S)$ is the free group with basis $S$. This is the universal property of the free group generated by $S$. But $F(S_1)$ is nothing else than the infinite cyclic group with one generator $1$.
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
$endgroup$
Let $S_1 = 1 $. Then you have a natural bijection between $mathbfSet(S_1,UG)$ and $UG$. Moreover, for any set $S$, you have a natural bijection between $mathbfSet(S,UG)$ and $mathbfGrp(F(S),G)$, where $F(S)$ is the free group with basis $S$. This is the universal property of the free group generated by $S$. But $F(S_1)$ is nothing else than the infinite cyclic group with one generator $1$.
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
answered Mar 27 at 22:48
Paul FrostPaul Frost
12.9k31035
12.9k31035
add a comment |
add a comment |
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$begingroup$
If two objects represent the same functor, they must be isomorphic.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 20:42