There are many bijective functions that map $\mathbb{R}$ to $(-1,1)$, in particular: $$f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$$
(Of course there are others, such as $g\left(x\right)=\frac{2}{\pi}\arctan(x)$, etc., but I like mine since $f\left(0\right)=0$ and $f^\prime\left(0\right)=1$.)
$f\left(x\right)$ is indeed a bijection, and it also preserves order, since $x_1\le x_2$ implies $f\left(x_1\right)\le f\left(x_2\right)$. I'll omit the proofs for the sake of wordiness. Incidentally, my conjecture is that all order-isomorphic functions that map $\mathbb{R}$ to $(-1,1)$ are sigmoid functions, but I digress.
My first question is about the nature of the isomorphism $f$. Obviously $f$ is order-preserving, but is it operation-preserving? Does there exist an operation $\#$ on $(-1,1)$ such that $f(x_1)\#f(x_2)=f(x_1+x_2)$? or $\ast$ such that $f(x_1)\ast f(x_2)=f(x_1\cdot x_2)$?
Also, my second question: In layman's terms, two sets that are "isomorphic" are "essentially the same." That is, you could treat both sets as the same set, except that maybe the elements have different names*. I don't see how this is possible, since $(-1,1)$ has a least upper bound, whereas $\mathbb{R}$ does not have a least upper bound. $\mathbb{R}$ and $(-1,1)$ don't "look" the same. All the other isomorphic sets I can think of "look" (structurally) the same.
*The sets $\left\{\text{one},\text{two},\text{three}\right\}$ and $\left\{\textit{uno},\textit{dos},\textit{tres}\right\}$ are order- and operation-isomorphic.
EDIT: I just discovered that the function $f(x)$ is exactly $\tanh(x)$.
