I'm looking for the domain of definition of the composite function of one function and its inverse.
Let $$H(x)= \frac{e^x -1}{e^x +1} $$ and its inverse $$G(x)=\ln{ \frac{x+1}{1-x}}$$
Is $-1 <x <1 $ the domain of definition of $H(G(x))$ ?
What is weird is that by going through the computer algebra tool Mathematica, I get $\mathbb{R}$ as the domain. I think the tool does a simplification and gets the identity function, and that explains why I get $\mathbb{R}$ as the domain. Am I right?
The graph of $$ \frac{x+1}{1-x}: $$
The graph of $H(x)$ in blue and $G(x)$ in orange:
This is one of those places in mathematics where the language gets a little muddy, and where authors often use words differently. From reading your question I can infer some things, but it would not be unreasonable to consult the source from which you are working.
Usually, we require three data when defining a function: (1) a domain, (2) a codomain, and (3) a (unique) pairing of each element in a the domain to some element of the codomain. However, as a practical matter, we often elide the an explicit statement of the domain and/or codomain, and take the domain to be the largest set in our "universe" for which the formula makes sense. This gives us a domain of definition.
In this case, it appears that you are working over $\mathbb{R}$. Thus both $G$ and $H$ are functions which act on subsets of the real numbers. The function $H$ is defined by the formula $$ H(x) = \frac{\mathrm{e}^x-1}{\mathrm{e}^x + 1}. $$ This formula gives a well-defined real number for any real number $x$, and so we say that the domain of definition for $H$ is $\mathbb{R}$.
The function $G$ is defined by the formula $$ G(x) = \ln\left( \frac{x+1}{1-x} \right). $$ This function does not give a well-defined real number for all real numbers $x$. Indeed, if $(x+1)/(x-1) \le 0$, then the logarithm is undefined. Thus the domain of definition for $G$ cannot contain any values of $x$ for which this fraction is nonpositive. Some quick algebra shows that $$ \frac{x+1}{x-1} > 0 \iff x \in (-1, 1). $$ For any such $x$, the formula defining $G$ gives a well-defined real number, and so the domain of definition for $G$ is $(-1,1)$.
Finally, given two function $f$ and $g$ defined by formulae, the domain of definition of the composition $f\circ g$ consists of all of the points in the domain of definition of $g$, minus those points that are not mapped into the domain of definition of $f$. That is, $$ D(f\circ g) = D(g) \setminus \{ x\in D(f) \mid g(x) \not\in D(f) \}. $$ For the given functions $G$ and $H$, $$ D(H\circ G) = D(G) \setminus \underbrace{\{ x\in D(G) \mid G(x) \not\in D(H) \}}_{\text{$=\varnothing$, see below}} = (-1,1)\setminus \varnothing = (-1,1).$$ The indicated term is the empty set, as there are no real values of $x$ (in the domain of $G$ or otherwise) which are not in the domain of $H$.
In other words, $H\circ G$ is only defined on the interval $(-1,1)$, and so this interval is the domain of definition of $H\circ G$. On this interval, it is the identity function, i.e. $(H\circ G)(x) = x$.
Observe that if we define a function $F$ by $F(x) = x$, then this function has $\mathbb{R}$ as its domain of definition, because the defining formula is well-defined for all real inputs. However, $F$ and $H\circ G$ are not the same function: $F$ is defined by a formula with its domain understood implicitly, while $H\circ G$ is defined by the composition of two functions with previously understood domains of definition. If you are using a computer algebra system (CAS) to compute the domain of the composition, it is very likely that the CAS is first simplifying the resulting formula for the composition, and then determining the domain. This highlights how careful you need to be when working with computers.