# Does every surjective morphism from an uncountable into a countable monoid admit a homomorphic right inverse function

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.

• 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. – Qiaochu Yuan May 9 '16 at 18:17
• @QiaochuYuan Thanks, I will change the title. – 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 $\mathbb{Q}^{\mathbb{N}}$).
• 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? – StefanH May 11 '16 at 17:51
• @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. – Qiaochu Yuan May 11 '16 at 17:53