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).

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 $K$.

  • 1
    $\begingroup$ Since $\pi_H$ is defined in the free monoid on H why would $\pi_H^{-1}$ be contained in the free monoid on the atoms? $\endgroup$ Commented Apr 1, 2017 at 17:13
  • $\begingroup$ My fault, it's a mistake: I want $\mathscr{Z}_H(x) := \pi_H^{-1}(x) \cap \mathscr{F}^\ast(\mathcal A(H))$ for every $x \in H$. Besides that, I think I can prove that the answer is in the affirmative. I will post it later. $\endgroup$ Commented Apr 1, 2017 at 17:30
  • $\begingroup$ For the record, the converse is true under mild assumptions, that are often met in applications (where K is typically a monoid of zero-sum sequences over an abelian group $G$ with support in a set $G_0\subseteq G$): (i) if $K$ is reduced (i.e., the only unit of $K$ is the identity) and atomic; (ii) if $K$ is atomic and $\varphi$ is a weak transfer homomorphism (viz., an essentially surjective equimorphism with the additional property that $\varphi^{-1}(K^\times)\subseteq H^\times$). $\endgroup$ Commented Apr 3, 2017 at 17:30

1 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.

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. []

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.


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