# Isomorphism from $\mathbb{R}$ to $(-1,1)$

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 $\tanh(x)$.

-
since it's a bijection I think you can just define y1 # y2 = f(f^{-1}(y1)+f^{-1}(y2)). –  user58512 Jan 26 '13 at 18:33
To address the second question, I wouldn't say the sets look the same. I would instead say that the algebraic structure looks the same. Your example is a nice one for many reasons, it also shows that completeness is not a topological or algebraic property, for example. –  Clayton Jan 26 '13 at 18:35
There is absolutely no need to create the tag [isomorphism]. If you think this tag is useful please start a meta discussion about it. –  Asaf Karagila Jan 26 '13 at 18:38
But it's not a helpful tag. 99.99% of questions about isomorphism will contain the word [isomorphism], and 99.99% of questions containing the word isomorphism have merits to having the tag [isomorphism]. This makes a very simple argument against using it as a tag. –  Asaf Karagila Jan 26 '13 at 18:54
I removed the isomorphism tag. To add to @Asaf's comment, to state something is an "isomorphism" is completely useless if you don't specify what algebraic structures you are interested in. –  Willie Wong Jan 28 '13 at 9:04

You just defined $\#$ and $*$ in terms of $f$. Namely, the operations: $$a\# b=f\left(f^{-1}(a)+f^{-1}(b)\right)\\a*b=f\left(f^{-1}(a)\cdot f^{-1}(b)\right)$$
Obviously $f$ respects these operations, and it's not easy to see that they are unique. Furthermore $(0,1)$ with $\#$ and $*$ forms a field, which is isomorphic to $\mathbb R$ via $f$. This is also known as transport of structure.
As for your second question, note that internally $(0,1)$ does not have an upper bound. If two structures are isomorphic this means that you can treat them as the same structure, up to replacing the elements. So only things which are internally definable from this structure should be relevant.
Note that as topological spaces $(0,1)$ and $\mathbb R$ are the same, but as metric spaces they are not isomorphic because $(0,1)$ has a finite width and $\mathbb R$ not. You can't use everything you know about the particular set, but rather the things which are preserved by the isomorphism.
In what sense does $(0,1)$ have "finite width"? –  Pedro Tamaroff Feb 3 '13 at 3:33