topological groups - Is there a "nice" discontinuous, bijective homomorphism $f: (mathbb{R},+) to (mathbb{R},+)$?

Consider $(\mathbb{R},+)$ as a topological group. Using the axiom of choice, we can construct a $\mathbb{Q}$-basis for $\mathbb{R}$ and using this basis, we can define a discontinuous, bijective homomorphism from $(\mathbb{R},+)$ to itself.

Is it possible to find such a homomorphism without using the axiom of choice?

Answer

The answer is no, we need some choice to construct such homomorphism, because it is consistent with ZF (without choice) that every function $\phi:\Bbb R\rightarrow\Bbb R$ satifying $\phi(x+y)=\phi(x)+\phi(y)$ is continuous. You can get far more details in this MO answer.

Here you can see a handful of collected facts about the nontrivial solutions, provided any exist.