If $H$ is essentially equimorphic to $K$, then is $H$ atomic only if so is $K$?

Question

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 $K$?

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 $K$.

Answer 1

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 that $K$ is Dedekind-finite (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.