Let $M$ be a an uncountable monoid (like $\mathbb R$ with addition or multiplication) and $N$ be a countable monoid (like $\mathbb N_0$, or $\mathbb Z$ with addition or multiplication). Further suppose we have a surjective homomorphism $\varphi : M \to N$ (like $(\mathbb R, \mbox{max})$ and $\mathbb (\mathbb Z, \mbox{max})$ with $\varphi(x) := \mbox{min}\{ k \in \mathbb Z : x \le k \}$, i.e. rounding up).

In this situation, does there always exists a homomorphism $\psi : N \to M$ such that $\varphi \circ \psi = \mbox{id}_N$.

For the terminology as used in the title, see wikipedia:surjections as epimorphisms.

  • 2
    $\begingroup$ Your title and body questions don't match: it's not true that the epimorphisms of monoids are the surjections. For example, the inclusion of $\mathbb{Z}_{\ge 0}$ into $\mathbb{Z}$ is an epimorphism but not a surjection. $\endgroup$ – Qiaochu Yuan May 9 '16 at 18:17
  • $\begingroup$ @QiaochuYuan Thanks, I will change the title. $\endgroup$ – StefanH May 11 '16 at 17:51

No. We can even write down a counterexample for abelian groups: take $M = \mathbb{R} \oplus \mathbb{Z}$ and take $N = \mathbb{Z}/2\mathbb{Z}$. If you really want $N$ to be countable as opposed to at most countable, take $N = \mathbb{Q} \oplus \mathbb{Z}/2\mathbb{Z}$ (there's a surjection $\mathbb{R} \to \mathbb{Q}$ assuming the axiom of choice; without choice replace $\mathbb{R}$ with $\mathbb{Q}^{\mathbb{N}}$).

More generally, in the setting of groups at least, look up the group extension problem.

  • $\begingroup$ For this construction with $\mathbb R$ we need a homomorphic surjection $\mathbb R \to \mathbb Q$, but the axiom of choice just guarantees there exists a surjective function? So what have I missed? $\endgroup$ – StefanH May 11 '16 at 17:51
  • 2
    $\begingroup$ @Stefan: using the axiom of choice you can show that $\mathbb{R}$ has a basis as a $\mathbb{Q}$-vector space, and using this basis you can write down a quotient map to $\mathbb{Q}$ by quotienting by every element of the basis except one. $\endgroup$ – Qiaochu Yuan May 11 '16 at 17:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.