Defining a map on a group 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.
 A: 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.
