Homomorphism between integers and $(-1,1)$ under any operation contains an irrational point

Question

I need to prove that

Any homomorphism $h$ between integers and $(-1,1)$ under any operation contains an irrational point, i.e. there exists a $k\in\mathbb Z$ such that $h(k)$ is irrational.

For example, consider the group $(-1,1)$ with the operation $a\oplus b=\dfrac{a+b}{1+ab}$. It is not hard to see that $h:\mathbb Z\to(-1,1):=k\mapsto \tanh(k)$ is a homomorphism from $\mathbb Z$ under regular addition to $(-1,1)$ under $\oplus$, and, clearly, $h(1)=\tanh(1)$ is irrational, so we have found an irrational point in the homomorphism.

Is it true for any operation and homomorphism?

I feel that the fact that $(-1,1)$ is bounded and the integers not may play a role in the argument, but I cannot see how, any help would be appreciated. Thanks.

Answer 1

It's not true. There is a one-to-one function $f_1$ from $\mathbb Q \cap (-1,1)$ onto $\mathbb Z$ (because they have the same cardinality). Similarly there is a one-to-one function $f_2$ from $(-1,1) \backslash \mathbb Q$ onto $\mathbb R \backslash \mathbb Z$. Put them together to get a one-to-one function $f$ from $(-1,1)$ onto $\mathbb R$ that maps $\mathbb Q \cap (-1,1) \to \mathbb Z$. Let the operation on $(-1,1)$ be $\oplus$ defined by $$ x \oplus y = f^{-1}(f(x) + f(y))$$ This makes $(-1,1)$ into a group and $f$ into a homomorphism (in fact an isomorphism). $h = f_1^{-1}$ is a homomorphism from $\mathbb Z$ into $(-1,1)$ whose image contains only rationals.

Answer 2

This is false. For any operation on $(-1,1)$ with a rational identity (like the example operation in your question, which has $0$ as an identity), there exists a homomorphism that sends all of $\mathbb Z$ to that rational identity. Hence, that homomorphism has no irrational points in its image.

Answer 3

As we say in mathematics, the answer is trivially NO because you allowed for a total set-theoretical freedom of selecting $h$. You can easily get even a strictly increasing $\ h\ $ which can be interpreted as a homomorhism which has rational values only:

Define bijections $\ H:\Bbb R\to(-1;1)\ $ and $\ F:(-1;1)\to\Bbb R\ $ -- each inverse to the other -- as follows,

$$ \forall_{x\in\Bbb R}\quad H(x)\ :=\ \frac x{1+|x|} $$

$$ \forall_{y\in(-1;1)}\quad F(y)\ :=\ \frac y{1-|y|} $$

Finally, define binary operation $\ \#\ $ in $\ (-1;1)\ $ by borrowing it from $\ \Bbb R$,

$$ \forall_{a\ b\,\in(-1;1)}\quad a\,\#\,b\,\ :=\,\ H(F(a) + F(b)) $$

Thus $\ \#\ $ is a group operation such that $\ H:\Bbb R\to(-1;1)\ $ and $\ F:(-1;1)\to\Bbb R\ $ are isomorphisms, each inverse to the other.

Finally, define $\ h:\Bbb Z\to(-1;1)\ $ as

$$ h\ :=\ H|\Bbb Z $$

Then $\ h\ $ is the required homomorphism (monomorphism) such that

$$ \forall_{n\in\Bbb Z}\quad h(n)\in\Bbb Q.$$

Great!