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David White
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Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z$\mathbb{Z}$ in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z$\mathbb{Z}$ is that it is a separator, i.e. for any two distinct pair of parallel arrows f,g:A --> B$f,g:A \rightarrow B$, there is at least one morphism x:Z --> A$x:\mathbb{Z} \rightarrow A$ such that fx \neq gx$f\circ x \neq g\circ x$. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperatorsseparators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z is that it is a separator, i.e. for any two distinct pair of parallel arrows f,g:A --> B, there is at least one morphism x:Z --> A such that fx \neq gx. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers $\mathbb{Z}$ in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of $\mathbb{Z}$ is that it is a separator, i.e. for any two distinct pair of parallel arrows $f,g:A \rightarrow B$, there is at least one morphism $x:\mathbb{Z} \rightarrow A$ such that $f\circ x \neq g\circ x$. Unfortunately, any free group satisfies this property. I have two questions:

What are the separators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

Fixed spelling.
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Qiaochu Yuan
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Seperators Separators in the Category of Groups

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z is that it is a seperatorseparator, i.e. for any two distinct pair of parallel arrows f,g:A --> B, there is at least one morphism x:Z --> A such that fx \neq gx. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

Seperators in the Category of Groups

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z is that it is a seperator, i.e. for any two distinct pair of parallel arrows f,g:A --> B, there is at least one morphism x:Z --> A such that fx \neq gx. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

Separators in the Category of Groups

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z is that it is a separator, i.e. for any two distinct pair of parallel arrows f,g:A --> B, there is at least one morphism x:Z --> A such that fx \neq gx. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?

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Steven Gubkin
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Seperators in the Category of Groups

Some of my friends and I were trying to discover a universal mapping property that characterizes the integers Z in the category of groups without referring to the underlying sets (So it is a no no to say it is the free group on a one element set). One of the big uses of Z is that it is a seperator, i.e. for any two distinct pair of parallel arrows f,g:A --> B, there is at least one morphism x:Z --> A such that fx \neq gx. Unfortunately, any free group satisfies this property. I have two questions:

What are the seperators in the category of groups (I think they will be just the free groups, but I have not proved this yet). Given this I think I can write down a universal property for Z which stays inside the category of groups.

Whether or not the above claim is correct, does anyone have a UMP that does the job?