Without axiom of choice, is there a surjective homomorfism $ (\mathbb{R},+)\rightarrow(\mathbb{Q},+) $ ?
Or some nontrivial homomorphism?
2026-04-02 19:22:18.1775157738
Group-homomorphism $ \mathbb{R\rightarrow\mathbb{Q}} $ without Axiom of choice
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No.
If every set of reals has the Baire property, then every group homomorphism from the reals into the rationals is continuous, and therefore trivial.
This assumption is consistent, as shown by Solovay. Even with Dependent Choice.