I will first state my question, and then give all the relevant definitions.



> 
> **Q.** Let $H$ and $K$ be monoids, and assume $H$ is essentially equimorphic to $K$. Is it true that $H$ is atomic only if so is $K$?
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*Edit (Apr 03, 2017).* I thought I could prove that, if $H$ is essentially equimorphic to $K$ and $K$ is atomic, then so is $H$, and I was stating my question in the form of an "if and only if". However, when I started typing my scribbles, I realized that my argument for the "if" part had a gap, and here is now a counterexample: Let $K$ be the group of rational numbers under addition, and $H$ the submonoid of $K$ consisting of the non-negative rational numbers. Then $K$ is atomic (as any other group), and the canonical embedding $\varphi: H \to K: x \mapsto x$ is an essentially surjective equimorphism (in particular, the set of atoms of $H$ and the set of atoms of $K$ are both empty, so conditions (E2) and (E3) in the definitions below are vacuously true). Yet, $H$ is *not* atomic (as any other divisible monoid which is not a group).




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**Basic dictionary**


We denote by $\mathscr{F}^\ast(\mathscr{U})$, for a fixed set $\mathscr{U}$, the free monoid with basis $\mathscr{U}$. We use the symbol $\ast$ for the operation of $\mathscr{F}^\ast(\mathscr{U})$, and we take $\|1\_{\mathscr{F}^\ast(\mathscr{U})}\|\_\mathscr{U} := 0$ and $\|z\_1 \* \cdots \* z\_n\|\_\mathscr{U} := n$ for all $z\_1, \ldots, z\_n \in \mathscr{U}$. Given $\mathfrak z \in \mathscr{F}^\ast(\mathcal A(H))$, we call $\|\mathfrak z\|\_\mathscr{U}$ the *length* of $\mathfrak z$.


With this in hand, let $H$ be a multiplicatively written monoid. We denote by $H^\times$ the *set of units* (or *invertible elements*) of $H$, by $\mathcal A(H)$ the *set of atoms* of $H$ (an element $a \in H$ is an atom if $a \notin H^\times$ and there do not exist $x, y \in H \setminus H^\times$ such that $a = xy$), by $\pi\_H$ the unique homomorphism $\mathscr{F}^\ast(H) \to H$ such that $\pi\_H(x) = x$ for every $x \in H$,
and by $\mathscr{C}\_H$ the smallest (monoid) congruence on $\mathscr{F}^\ast(\mathcal A(H))$ determined by the following condition:


* If $\mathfrak a = a\_1 \ast \cdots \ast a\_m$ and $\mathfrak b = b\_1 \ast \cdots \ast b\_n$ are, respectively, non-empty $\mathcal A(H)$-words of length $m$ and $n$, then $(\mathfrak a, \mathfrak b) \in \mathscr{C}\_H$ if and only if $\pi\_H(\mathfrak a) = \pi\_H(\mathfrak b)$, $m = n$, and $a\_1 \simeq\_H b\_{\sigma(1)}, \ldots, a\_n \simeq\_H b\_{\sigma(n)}$ for some $\sigma \in \mathfrak S\_n$.


Here, $\mathfrak S\_n$ is the group of permutations of $[\![ 1, n ]\!]$, and $x \simeq\_H y$, for $x, y \in H$, means that $y \in H^\times x H^\times$ (viz., $x$ and $y$ are *associate*).
Moreover, we define, for every $x \in H$, 
$$
\mathscr{Z}\_H(x) := \pi\_H^{-1}(x) \cap \mathscr{F}^\ast(\mathcal A(H)) \subseteq \mathscr{F}^\ast(\mathcal A(H))
$$ 
(the *set of factorizations* of $x$) and 
$$\mathsf L\_H(x) := \{\|\mathfrak a\|\_H: \mathfrak a \in \mathscr{Z}\_H(x)\}$$
(the *set of lengths* of $x$). We call $H$ *atomic* if $\mathsf L\_H(x) \ne \emptyset$ for all $x \in H \setminus H^\times$.


Next, let $H$ and $K$ be multiplicatively written monoids, and let $\varphi$ be a homomorphism $H \to K$. We write $\varphi^\ast$ for the unique homomorphism $\mathscr{F}^\ast(H) \to \mathscr{F}^\ast(K)$ such that $\varphi^\ast(x) = \varphi(x)$ for all $x \in H$, and we say that $\varphi$ is *essentially surjective* if $K = K^\times \varphi(H) K^\times$ (this is actually an instance of the notion of essentially surjective functor in category theory), and an *equimorphism* (*from $H$ to $K$*) if the following hold:


1. $\varphi(x) = 1\_K$ for some $x \in H$ only if $x \in H^\times$, that is, $\varphi^{-1}(1\_K) \subseteq H^\times$.
2. $\varphi$ is *atom-preserving*, i.e., $\varphi(a) \in \mathcal A(K)$ for all $a \in \mathcal A(H)$.
3. If $x \in H \setminus \{1\_H\}$ and $\mathfrak b \in \mathscr{Z}\_K(\varphi(x)) \ne \emptyset$, then $(\mathfrak b, \varphi^\ast(\mathfrak a)) \in \mathscr{C}\_K$ for some $\mathfrak{a} \in \mathscr{Z}\_H(x)$.


Lastly, we say that $H$ is *essentially equimorphic* to $K$ if there exists an essentially surjective equimorphism from $H$ to $K$.



If $H$ is essentially equimorphic to $K$, then is $H$ atomic only if so is $K$?
Sorry for answering my own question: It's just that the extra effort you put in making things clear for others, does often make things clearer for yourself, in the first place.




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Something stronger and more general is true. Indeed, we have the following:


**Proposition.** Let $\varphi: H \to K$ be an atom-preserving, essentially surjective homomorphism, and suppose that $H$ is atomic. Then $\varphi^{-1}(K^\times) = H^\times$ and $K$ is atomic too.


*Proof.* Since $\varphi$ is a homomorphism, we have $\varphi(H^\times) \subseteq K^\times$. Therefore, if $H$ is a group, we are done, because $\varphi$ being essentially surjective yields $K = K^\times \varphi(H) K^\times \subseteq K^\times$, with the result that also $K$ is a group, and groups are vacuously atomic. 


Consequently, we assume for the remainder of the proof that $\mathcal A(H)$ is non-empty. Then so is $\mathcal A(K)$, since $\emptyset \ne \varphi(\mathcal A(H)) \subseteq \mathcal A(K)$ by the hypothesis that $\varphi$ is atom-preserving. It follows from this and a [cute observation of Benjamin Steinberg](https://mathoverflow.net/a/261870/16537) that $K$ is [Dedekind-finite](https://mathoverflow.net/questions/261982/) (equivalently, $yz \in K^\times$, for some $y, z \in K$, if and only if $y, z \in K^\times$).


Suppose for a contradiction that $\varphi^{-1}(K^\times) \neq H^\times$. Then, there exists $x \in H \setminus H^\times$ such that $\varphi(x) \in K^\times$. But $H$ is atomic, so $x = a\_1 \cdots a\_n$ for some $a\_1, \ldots, a\_n \in \mathcal A(H)$, which gives in turn that $\varphi(a\_1) \cdots \varphi(a\_n) \in K^\times$. This is, however, impossible: $\varphi$ being atom-preserving implies that $\varphi(a\_1), \ldots, \varphi(a\_n)$ are atoms of $K$ (in particular, they are *not* invertible), and hence their product cannot be a unit, as we have seen that $K$ is Dedekind-finite. 


Thus, $\varphi^{-1}(K^\times) = H^\times$ and we are left to prove that $K$ is atomic. To this end, pick $y \in K \setminus K^\times$. Then $y \simeq\_K \varphi(x)$ for some $x \in H$, by the hypothesis that $\varphi$ is essentially surjective. But $x$ cannot be a unit of $H$, so $x = a\_1 \cdots a\_n$ for some atoms $a\_1, \ldots, a\_n \in H$, and hence $\varphi(x) = b\_1 \cdots b\_n$ for some $b\_1, \ldots, b\_n \in \mathcal A(K)$ (I'm just reusing the argument from the previous paragraph). To wit, $y$ is a non-empty product of atoms of $K$, when considering that the result of pre- and post-multiplying an atom by a unit is still an atom (in any monoid). This finishes the proof. []




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Incidentally, an atom-preserving, essentially surjective homomorphism $\varphi: H \to K$ for which $\varphi^{-1}(K^\times) = H^\times$ is called a *weak transfer homomorphism*. So, the above proposition shows, in particular, that an atom-preserving, essentially surjective homomorphism whose source is atomic, is necessarily a weak transfer homomorphism. 


