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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?

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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.

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