Without axiom of choice, is there a surjective homomorfism $ (\mathbb{R},+)\rightarrow(\mathbb{Q},+) $ ?

Or some nontrivial homomorphism?

  • 2
    $\begingroup$ Every nontrivial homomorphism is surjective. $\endgroup$
    – lhf
    Jan 4, 2018 at 12:28

1 Answer 1



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.

  • $\begingroup$ Could you expand a bit on how the Baire property implies that group homomorphisms are continuous ? I'm guessing this has something to do with the homogeneity of the kernel; but I can't really see $\endgroup$ Jan 4, 2018 at 13:43
  • $\begingroup$ @Max: It's a nontrivial theorem, that every Baire measurable homomorphism is continuous (the same is true for Lebesgue measurable). The theorem is originally due to Banach from the 1920s, if my memory serves me right. You can find information in some notes on my website. I will try to edit more references later. $\endgroup$
    – Asaf Karagila
    Jan 4, 2018 at 14:43
  • $\begingroup$ Oh ok, the way you put it I thought it was easy ! I'll take a look at your website $\endgroup$ Jan 4, 2018 at 14:44

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