Is division by two continuous in topological groups? 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. 
 A: 
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.
A: The answer is no, even assuming the group to be locally compact.
Let $I$ be an infinite set. Endow $\mathbf{Q}_2^I$ with the topology making $\mathbf{Z}_2^I$ a compact open subgroup (that is, the topology induced by inclusion into $\mathbf{Q}_2^I\times (\mathbf{Q}_2/\mathbf{Z}_2)^I$, the first factor being endowed with the product topology and the second factor being endowed with the discrete topology).
Then multiplication by $1/2$ is not continuous. Indeed, it maps the compact subgroup $\mathbf{Z}_2^I$ onto $(\frac12\mathbf{Z}_2)^I$, which is not compact since its quotient $(\frac12\mathbf{Z}_2)^I/\mathbf{Z}_2^I$ is infinite discrete.

However the answer is yes for a $\sigma$-compact locally compact abelian group. Because in this case every surjective homomorphism is a quotient map, and hence multiplication by 2 being a bijective quotient map, has to be open.
A: 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.  
