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Why is $mathbbZ$ not an inital object of Gr or AB?
definition of a free object in a categoryWhy is an empty set not a terminal object in categories $mathsfTop$ and $mathsfSets$?Equivalence of group objects in set and groups as one object categories.Initial and Final Objects in a CategoryArrow From an Initial Object Does Not Disturb Commutativity of the DiagramWhen the empty family of arrows to an object is epimorphic, that object must be initial?Zero objects in preadditive categoriesWhat does “$textHom(space, G)$takes values in Groups” mean?Show that the terminal object is unique up to a unique isomorphismA weaker notion of exponential object?
$begingroup$
Why is $mathbbZ$ not an inital object of GR or AB?
Claim 1: for every group $G$ there exists a groups morphism from $mathbbZ$ to $G$.
PF: Let $f:mathbbZ rightarrow G$ be given by: $f(n) = n*1_G$.
Clearly $f(1) = 1_G$. Now $f(n+m) = (n+m)*1_G = n*1_G + m*1_G$.
Hence $f$ is a groups hom.
Claim 2: $f$ is unique.
PF: This follows from that $mathbbZ$ is generated by $1$ and hence every group hom starting from $mathbbZ$ is completely determined by the image of $1$. And this image has to be $1_G$ by definition of group hom.
Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.
abstract-algebra group-theory category-theory integers
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
$endgroup$
add a comment |
$begingroup$
Why is $mathbbZ$ not an inital object of GR or AB?
Claim 1: for every group $G$ there exists a groups morphism from $mathbbZ$ to $G$.
PF: Let $f:mathbbZ rightarrow G$ be given by: $f(n) = n*1_G$.
Clearly $f(1) = 1_G$. Now $f(n+m) = (n+m)*1_G = n*1_G + m*1_G$.
Hence $f$ is a groups hom.
Claim 2: $f$ is unique.
PF: This follows from that $mathbbZ$ is generated by $1$ and hence every group hom starting from $mathbbZ$ is completely determined by the image of $1$. And this image has to be $1_G$ by definition of group hom.
Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.
abstract-algebra group-theory category-theory integers
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
$endgroup$
1
$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
2
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57
add a comment |
$begingroup$
Why is $mathbbZ$ not an inital object of GR or AB?
Claim 1: for every group $G$ there exists a groups morphism from $mathbbZ$ to $G$.
PF: Let $f:mathbbZ rightarrow G$ be given by: $f(n) = n*1_G$.
Clearly $f(1) = 1_G$. Now $f(n+m) = (n+m)*1_G = n*1_G + m*1_G$.
Hence $f$ is a groups hom.
Claim 2: $f$ is unique.
PF: This follows from that $mathbbZ$ is generated by $1$ and hence every group hom starting from $mathbbZ$ is completely determined by the image of $1$. And this image has to be $1_G$ by definition of group hom.
Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.
abstract-algebra group-theory category-theory integers
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
$endgroup$
Why is $mathbbZ$ not an inital object of GR or AB?
Claim 1: for every group $G$ there exists a groups morphism from $mathbbZ$ to $G$.
PF: Let $f:mathbbZ rightarrow G$ be given by: $f(n) = n*1_G$.
Clearly $f(1) = 1_G$. Now $f(n+m) = (n+m)*1_G = n*1_G + m*1_G$.
Hence $f$ is a groups hom.
Claim 2: $f$ is unique.
PF: This follows from that $mathbbZ$ is generated by $1$ and hence every group hom starting from $mathbbZ$ is completely determined by the image of $1$. And this image has to be $1_G$ by definition of group hom.
Now something is fishy here, because the trivial group is supposed to be the initial object of GR. And initial objects, when they exsist are unique.
abstract-algebra group-theory category-theory integers
abstract-algebra group-theory category-theory integers
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
asked Mar 21 at 13:51
Jens WagemakerJens Wagemaker
581312
581312
1
$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
2
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57
add a comment |
1
$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
2
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57
1
1
$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
2
2
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You've mixed up the additive and multiplicative identity of $mathbbZ$. $(mathbbZ,+)$ is a group with identity $0$. A group homomorphism from $mathbbZ$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $mathbbZ$ to a given group, in fact, mapping $1 in mathbbZ$ to any $g in G$ defines a homomorphism.
The initial and final objects of $mathbfGr$ and $mathbfAb$ are the trivial group $0$.
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
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$begingroup$
You've mixed up the additive and multiplicative identity of $mathbbZ$. $(mathbbZ,+)$ is a group with identity $0$. A group homomorphism from $mathbbZ$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $mathbbZ$ to a given group, in fact, mapping $1 in mathbbZ$ to any $g in G$ defines a homomorphism.
The initial and final objects of $mathbfGr$ and $mathbfAb$ are the trivial group $0$.
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
$begingroup$
You've mixed up the additive and multiplicative identity of $mathbbZ$. $(mathbbZ,+)$ is a group with identity $0$. A group homomorphism from $mathbbZ$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $mathbbZ$ to a given group, in fact, mapping $1 in mathbbZ$ to any $g in G$ defines a homomorphism.
The initial and final objects of $mathbfGr$ and $mathbfAb$ are the trivial group $0$.
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
add a comment |
$begingroup$
You've mixed up the additive and multiplicative identity of $mathbbZ$. $(mathbbZ,+)$ is a group with identity $0$. A group homomorphism from $mathbbZ$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $mathbbZ$ to a given group, in fact, mapping $1 in mathbbZ$ to any $g in G$ defines a homomorphism.
The initial and final objects of $mathbfGr$ and $mathbfAb$ are the trivial group $0$.
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
$endgroup$
You've mixed up the additive and multiplicative identity of $mathbbZ$. $(mathbbZ,+)$ is a group with identity $0$. A group homomorphism from $mathbbZ$ must take the additive identity $0$ to $1_G$, but this does not determine the homomorphism. There are many morphisms from $mathbbZ$ to a given group, in fact, mapping $1 in mathbbZ$ to any $g in G$ defines a homomorphism.
The initial and final objects of $mathbfGr$ and $mathbfAb$ are the trivial group $0$.
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
answered Mar 21 at 13:57
Joshua MundingerJoshua Mundinger
2,9171028
2,9171028
add a comment |
add a comment |
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$begingroup$
At some point you write $n ast 1_G + m ast 1_G$. What does this mean? If $G$ is a group, it carries only one operation. Did you mean to talk about (commutative) rings instead?
$endgroup$
– Bib-lost
Mar 21 at 13:56
2
$begingroup$
$mathbbZ$ is a group for the addition, not for the multiplication. Hence $1$ is not the neutral element and does not necessarily have to be mapped to $1_G$.
$endgroup$
– TastyRomeo
Mar 21 at 13:57