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In defining a map (homomorphism or not) on a group $G$, must one check well-definedness for both positive and negative elements separately? Or are there circumstances which allow one to specify the rule for a generic element of $G$ (of course not counting the trivial constant maps)?

EDIT: My question is phrased rather poorly. Let's take an example. Suppose I want to extend $2^{-}:\mathbb{N}\rightarrow\mathbb{R}_{\geq 0}$ to $\mathbb{Z}$. Can I just say that $2^{-}:\mathbb{Z}\rightarrow\mathbb{R}_{\geq 0}$ to $\mathbb{Z}$ is defined by the same rule that specifies the map from $\mathbb{N}$? Of course not: I have to specify the rule separately for the negative elements (am I always bound to send $-n$ to $1/2^{n}$?). I just wanted to know whether there exist general considerations that demand for a map to be specified in this manner.

spring
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    What do you mean by "positive and negative elements"? – Somos Dec 14 '19 at 21:06
  • I mean elements and their inverses: e.g., $5$ and $-5$ for integers – spring Dec 14 '19 at 21:07
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    Why not just prove it for all elements without restriction? Can you please give an example where you are unsure? – Somos Dec 14 '19 at 21:08
  • Is it always possible to define a map from a group by specifying the rule for all elements? (e.g. extending multiplication by $4$ from natural to integers) – spring Dec 14 '19 at 21:10
  • What is wrong with $n \mapsto 4n$ defined for all integers? – Somos Dec 14 '19 at 21:12
  • If you define a map, then all is fine. You define a map $f\colon G\to X$ by specifying a specific element $f(g)\in X$ for every possible $g\in G$. – Hagen von Eitzen Dec 14 '19 at 21:12
  • @Somos But do you mean to say that the map $2^{-}:\mathbb{Z}\rightarrow\mathbb{R}{\geq 0}$ (base 2 exponential function) is automatically specified when we specify $2^{-}:\mathbb{N}\rightarrow\mathbb{R}{\geq 0}$? – spring Dec 14 '19 at 21:18
  • You must define it for all elements of the group. As far as “well-definedness”, see these comments – Arturo Magidin Dec 14 '19 at 21:20
  • If you don't make any assumption on the map there is no way to extend automatically a function from $\mathbb{N}$ to $\mathbb{Z}$. On the other hand if your map is a monoid morphism into a group, then it extends uniquely to $\mathbb{Z}$. – Captain Lama Dec 14 '19 at 21:32

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Your clarified question is something like this. Suppose we have a group $G$ which has a set of free generators $F$ and a group $H$. Then any function $\,f:F\to H$ can be extended uniquely in the natural way to a homomorphism function $\,f^*:G\to H$ so that they agree on $F$ by the universal property of the free group. However, if we do not require the function to be a homomorphism, then we need to define it on the entire domain because that is what it means to define a function. Thus, $\,2^{-}:\mathbb{Z}\rightarrow\mathbb{R}_{\geq 0}\,$

Please refer to the Wikipedia article free group for more details about this.

However, what might be confusing you is a common abuse of notation. That is, by using an expression of the form $\, 2^n \,$ without specifying the domain of definition. In this and similar cases, the expression does not define a function because the domain is not explicitly given. A function requires both a domain and a rule giving values of the function in its domain. In other words $\,f(x):=2^x\,$ does not define a function unless its domain is explicitly (or much more likely implicitly) given. Thus, $\,2^{-}:\mathbb{Z}\rightarrow\mathbb{R}_{\geq 0} \,$ is okay since the domain is explicitly given and it is implictly assumed that $\,2^x\,$ has a well defined meaning.

Somos
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