Existence of injective homomorphism
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I'm interested in knowing if $operatorname{Hom}(-,S^1)$ is exact, understanding $S^1$ as the group of modulo $1$ complex numbers, is right exact on the category of finite abelian groups, because that would allow me to proof that every subgroup of such group $G$ is isomorphic to a quotient of $G$.
Right now I'm trying to show that it transform injections in surjections. Let $H,G$ finite abelian groups and $h:Hto G$ injective homomorphisms. I'll write $h^*$ for $operatorname{Hom}(h,S^1)$. To show that $h^*$ is surjective I take two homomorphisms $j: Fto H$ and $k:Fto H$ such that $j^*h^*=k^*h^*$. Let's apply this to $zin operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)Rightarrow zcirc hcirc j=zcirc hcirc k$.
If I could find an injective morphism $z$ then I would have $hcirc j=hcirc k$, which by injectivity of $h$ would result in $j=k$. I know in addition that $operatorname{Hom}(F,S^1)cong F$, so maybe I would just have to take the imagen of the element $(1,dots, 1)$ (given by the structure theorem) under this isomorphism. Since this isomorphism is provided by the fact that $mathbb{Z}_pcongmathbb{Z}_p^*$ identifying an element in $mathbb{Z}_p$ with the $p$-th root to which $1$ is mapped, $(1,dots, 1)^*$ would send the $i$-th $1$ to the first $p_i$-th root. Since each $p_i$ is different and they are primes, there are no coincident roots for different elements.
Is what I'm trying to do possible? I'd be thankful if someone could also answer to my very first question, which is: is $operatorname{Hom}(-,S^1)$ right exact in this context?
category-theory abelian-groups hom-functor
add a comment |
up vote
6
down vote
favorite
I'm interested in knowing if $operatorname{Hom}(-,S^1)$ is exact, understanding $S^1$ as the group of modulo $1$ complex numbers, is right exact on the category of finite abelian groups, because that would allow me to proof that every subgroup of such group $G$ is isomorphic to a quotient of $G$.
Right now I'm trying to show that it transform injections in surjections. Let $H,G$ finite abelian groups and $h:Hto G$ injective homomorphisms. I'll write $h^*$ for $operatorname{Hom}(h,S^1)$. To show that $h^*$ is surjective I take two homomorphisms $j: Fto H$ and $k:Fto H$ such that $j^*h^*=k^*h^*$. Let's apply this to $zin operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)Rightarrow zcirc hcirc j=zcirc hcirc k$.
If I could find an injective morphism $z$ then I would have $hcirc j=hcirc k$, which by injectivity of $h$ would result in $j=k$. I know in addition that $operatorname{Hom}(F,S^1)cong F$, so maybe I would just have to take the imagen of the element $(1,dots, 1)$ (given by the structure theorem) under this isomorphism. Since this isomorphism is provided by the fact that $mathbb{Z}_pcongmathbb{Z}_p^*$ identifying an element in $mathbb{Z}_p$ with the $p$-th root to which $1$ is mapped, $(1,dots, 1)^*$ would send the $i$-th $1$ to the first $p_i$-th root. Since each $p_i$ is different and they are primes, there are no coincident roots for different elements.
Is what I'm trying to do possible? I'd be thankful if someone could also answer to my very first question, which is: is $operatorname{Hom}(-,S^1)$ right exact in this context?
category-theory abelian-groups hom-functor
1
Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
@egreg yes, it is.
– Javi
Nov 17 at 14:52
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I'm interested in knowing if $operatorname{Hom}(-,S^1)$ is exact, understanding $S^1$ as the group of modulo $1$ complex numbers, is right exact on the category of finite abelian groups, because that would allow me to proof that every subgroup of such group $G$ is isomorphic to a quotient of $G$.
Right now I'm trying to show that it transform injections in surjections. Let $H,G$ finite abelian groups and $h:Hto G$ injective homomorphisms. I'll write $h^*$ for $operatorname{Hom}(h,S^1)$. To show that $h^*$ is surjective I take two homomorphisms $j: Fto H$ and $k:Fto H$ such that $j^*h^*=k^*h^*$. Let's apply this to $zin operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)Rightarrow zcirc hcirc j=zcirc hcirc k$.
If I could find an injective morphism $z$ then I would have $hcirc j=hcirc k$, which by injectivity of $h$ would result in $j=k$. I know in addition that $operatorname{Hom}(F,S^1)cong F$, so maybe I would just have to take the imagen of the element $(1,dots, 1)$ (given by the structure theorem) under this isomorphism. Since this isomorphism is provided by the fact that $mathbb{Z}_pcongmathbb{Z}_p^*$ identifying an element in $mathbb{Z}_p$ with the $p$-th root to which $1$ is mapped, $(1,dots, 1)^*$ would send the $i$-th $1$ to the first $p_i$-th root. Since each $p_i$ is different and they are primes, there are no coincident roots for different elements.
Is what I'm trying to do possible? I'd be thankful if someone could also answer to my very first question, which is: is $operatorname{Hom}(-,S^1)$ right exact in this context?
category-theory abelian-groups hom-functor
I'm interested in knowing if $operatorname{Hom}(-,S^1)$ is exact, understanding $S^1$ as the group of modulo $1$ complex numbers, is right exact on the category of finite abelian groups, because that would allow me to proof that every subgroup of such group $G$ is isomorphic to a quotient of $G$.
Right now I'm trying to show that it transform injections in surjections. Let $H,G$ finite abelian groups and $h:Hto G$ injective homomorphisms. I'll write $h^*$ for $operatorname{Hom}(h,S^1)$. To show that $h^*$ is surjective I take two homomorphisms $j: Fto H$ and $k:Fto H$ such that $j^*h^*=k^*h^*$. Let's apply this to $zin operatorname{Hom}(F,S^1)$. We get $j^*h^*(z)=k^*h^*(z)Rightarrow zcirc hcirc j=zcirc hcirc k$.
If I could find an injective morphism $z$ then I would have $hcirc j=hcirc k$, which by injectivity of $h$ would result in $j=k$. I know in addition that $operatorname{Hom}(F,S^1)cong F$, so maybe I would just have to take the imagen of the element $(1,dots, 1)$ (given by the structure theorem) under this isomorphism. Since this isomorphism is provided by the fact that $mathbb{Z}_pcongmathbb{Z}_p^*$ identifying an element in $mathbb{Z}_p$ with the $p$-th root to which $1$ is mapped, $(1,dots, 1)^*$ would send the $i$-th $1$ to the first $p_i$-th root. Since each $p_i$ is different and they are primes, there are no coincident roots for different elements.
Is what I'm trying to do possible? I'd be thankful if someone could also answer to my very first question, which is: is $operatorname{Hom}(-,S^1)$ right exact in this context?
category-theory abelian-groups hom-functor
category-theory abelian-groups hom-functor
edited 8 hours ago
asked Nov 17 at 14:28
Javi
2,4042725
2,4042725
1
Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
@egreg yes, it is.
– Javi
Nov 17 at 14:52
add a comment |
1
Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
@egreg yes, it is.
– Javi
Nov 17 at 14:52
1
1
Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
@egreg yes, it is.
– Javi
Nov 17 at 14:52
@egreg yes, it is.
– Javi
Nov 17 at 14:52
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $operatorname{Hom}({-},S^1)$ is exact.
A group $D$ (written additively) is divisible if, for every $xin D$ and every integer $nne0$, there is $yin G$ such that $x=ny$.
A group $E$ is injective if the functor $operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).
Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $nmathbb{Z}tomathbb{Z}$. Since $E$ is injective, the induced map $operatorname{Hom}(mathbb{Z},E)tooperatorname{Hom}(nmathbb{Z},E)$ is surjective.
Take $xin E$ and define $fcolon nmathbb{Z}to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $nmathbb{Z}$ of a homomorphism $gcolonmathbb{Z}to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.
The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $fcolon Ito E$, there exists $yin E$ such that $f(r)=ry$, for every $rin I$.
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
add a comment |
up vote
2
down vote
I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:Fto S^1$ to be the homomorphism that sends each generator $gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $operatorname{Hom}({-},S^1)$ is exact.
A group $D$ (written additively) is divisible if, for every $xin D$ and every integer $nne0$, there is $yin G$ such that $x=ny$.
A group $E$ is injective if the functor $operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).
Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $nmathbb{Z}tomathbb{Z}$. Since $E$ is injective, the induced map $operatorname{Hom}(mathbb{Z},E)tooperatorname{Hom}(nmathbb{Z},E)$ is surjective.
Take $xin E$ and define $fcolon nmathbb{Z}to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $nmathbb{Z}$ of a homomorphism $gcolonmathbb{Z}to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.
The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $fcolon Ito E$, there exists $yin E$ such that $f(r)=ry$, for every $rin I$.
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
add a comment |
up vote
4
down vote
accepted
The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $operatorname{Hom}({-},S^1)$ is exact.
A group $D$ (written additively) is divisible if, for every $xin D$ and every integer $nne0$, there is $yin G$ such that $x=ny$.
A group $E$ is injective if the functor $operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).
Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $nmathbb{Z}tomathbb{Z}$. Since $E$ is injective, the induced map $operatorname{Hom}(mathbb{Z},E)tooperatorname{Hom}(nmathbb{Z},E)$ is surjective.
Take $xin E$ and define $fcolon nmathbb{Z}to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $nmathbb{Z}$ of a homomorphism $gcolonmathbb{Z}to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.
The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $fcolon Ito E$, there exists $yin E$ such that $f(r)=ry$, for every $rin I$.
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $operatorname{Hom}({-},S^1)$ is exact.
A group $D$ (written additively) is divisible if, for every $xin D$ and every integer $nne0$, there is $yin G$ such that $x=ny$.
A group $E$ is injective if the functor $operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).
Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $nmathbb{Z}tomathbb{Z}$. Since $E$ is injective, the induced map $operatorname{Hom}(mathbb{Z},E)tooperatorname{Hom}(nmathbb{Z},E)$ is surjective.
Take $xin E$ and define $fcolon nmathbb{Z}to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $nmathbb{Z}$ of a homomorphism $gcolonmathbb{Z}to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.
The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $fcolon Ito E$, there exists $yin E$ such that $f(r)=ry$, for every $rin I$.
The group $S^1$ of complex numbers of modulo $1$ is divisible, so it is an injective object in the category of abelian group, which is tantamount to saying that the functor $operatorname{Hom}({-},S^1)$ is exact.
A group $D$ (written additively) is divisible if, for every $xin D$ and every integer $nne0$, there is $yin G$ such that $x=ny$.
A group $E$ is injective if the functor $operatorname{Hom}(-,E)$ is right exact (or exact, since it is left exact for every group).
Being divisible is a necessary condition for being injective. Indeed, let $E$ be injective and consider the injection $nmathbb{Z}tomathbb{Z}$. Since $E$ is injective, the induced map $operatorname{Hom}(mathbb{Z},E)tooperatorname{Hom}(nmathbb{Z},E)$ is surjective.
Take $xin E$ and define $fcolon nmathbb{Z}to E$ by $f(nk)=kx$. Then, by assumption, this map is the restriction to $nmathbb{Z}$ of a homomorphism $gcolonmathbb{Z}to E$. Then $x=f(n)=g(n)=ng(1)$, so taking $y=g(1)$ we have provided the element $y$ such that $x=ny$.
The converse is known as Baer's criterion: a right module $E$ over the ring $R$ is injective if and only if, for every right ideal $I$ of $R$ and every homomorphism $fcolon Ito E$, there exists $yin E$ such that $f(r)=ry$, for every $rin I$.
edited Nov 17 at 15:12
answered Nov 17 at 14:59
egreg
175k1383198
175k1383198
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
add a comment |
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
I don't know much of injective objects, if you could provide a reference of being divisible implies injective I would accept your answer.
– Javi
Nov 17 at 15:01
add a comment |
up vote
2
down vote
I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:Fto S^1$ to be the homomorphism that sends each generator $gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
add a comment |
up vote
2
down vote
I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:Fto S^1$ to be the homomorphism that sends each generator $gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
add a comment |
up vote
2
down vote
up vote
2
down vote
I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:Fto S^1$ to be the homomorphism that sends each generator $gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.
I think you can do this directly, since you are working in the category of finite abelian groups, because in this case, $F=bigoplus^n_{i=1} Z_{q_i}$, where the $q_i$ are powers of primes. So, you can take $z:Fto S^1$ to be the homomorphism that sends each generator $gamma_i$ of $Z_{q_i}$ to a particular $q_i$-th root of unity.
edited Nov 17 at 16:44
answered Nov 17 at 16:18
Matematleta
9,4992918
9,4992918
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
add a comment |
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
1
1
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
Thanks, I needed confirmation of that :)
– Javi
Nov 17 at 23:24
add a comment |
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Is $S^1$ the group of complex numbers with modulo $1$?
– egreg
Nov 17 at 14:51
@egreg yes, it is.
– Javi
Nov 17 at 14:52