Is division by two continuous in topological groups?

Question

Assume that $(G, +)$ is a Hausdorff topological abelian group which is uniquely divisible by two, i.e. the function $x \to 2x = x+ x$ is a bijection. Clearly, it is also continuous. My question is if this is an open mapping. Without unique division by two this needs not to be true, but I can't find an example if division by two is uniquely performable.

I have just spotted that the previous title was misleading, probably changed by another user during the edition. I wanted to ask if division by two is continuous, not open.

Answer 1

It seems the following.

The answer is positive. Let $d:G\to G$ be the division by two, $U\subset G$ be an arbitrary open set, and $x\in U$. Then $d(x)+d(x)=x\in U$. Since the addition on the group $G$ is continuous, there exists an open neighborhood $V$ of the point $d(x)$ such that $V+V\subset U$. Since the division by two is unique in the group $G$, $d(U)\supset V$. So the map $d$ is open.

PS. It is easy to check that $d$ is a homomorphism and that $G$ is uniquely divisible by two iff $G$ has no elements of order 2.

Answer 2

I have just spotted that the previous title was misleading, probably changed by another user during the edition. I wanted to ask if division by two is continuous, not open.

The answer, in general, is negative. Let $d:G\to G$ be the division by two, and $U\subset G$ be an arbitrary open set. Then $$d^{-1}(U)=\{d^{-1}(x):x\in U\}=\{x+x:x\in U\}.$$

The latter set is not necessarily open in every topological group $G$ which is uniquely divisible by two. For instance, pick a prime number $p>2$ and endow the additive group $\Bbb Q$ of rational numbers by a group topology with the family $\mathcal B=\{U_n\}$ of its open neighborhoods of the zero, where $U_n=p^n\Bbb Z$ for each natural $n$. Then $\Bbb Z$ is an open neighborhood of the zero in $\Bbb Q$, but $d^{-1}(\Bbb Z)=2\Bbb Z$ is not open, because for each $n$ a base open neighborhood $U_n$ of the zero contains an odd number.