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Consider two algebras $\mathscr{A}=\langle A,O_i\rangle$ and $\mathscr{B}=\langle B,P_i\rangle$ with the same type. Suppose that there is a homomorphism from $A$ onto $B$, and a homomorphism from $B$ onto $A$. Clearly (edit: not really), $|A|=|B|$. It is easy to prove that if the algebras are finite, they must be isomorphic, but can we conclude the same thing for infinite algebras? I can find a counterexample for the same question about relational structures (sets equipped with relations) but my counterexample uses a one-many relation (the structures are $\big\langle\mathbb{N},\{\langle1,1\rangle,\langle1,2\rangle\}\big\rangle$ and $\big\langle\mathbb{N},\{\langle1,1\rangle\}\big\rangle$).

Navid
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    Related: https://math.stackexchange.com/questions/176972/is-there-a-cantor-schroder-bernstein-statement-about-surjective-maps – Eran Mar 16 '21 at 12:55
  • @Eran, not really. – Navid Mar 16 '21 at 13:05
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    It certainly doesn't answer your question, but it does explain that your use of "Clearly" requires the assumption of some form of choice. – Eran Mar 16 '21 at 13:13
  • @Eran Oh! I didn't pay attention to that. Thanks! – Navid Mar 16 '21 at 15:31
  • @Navid How do you define a homomorphism of relational structures $(A, R)$ and $(B, S)$? According to this source, $\phi: A\to B$ is a homomorphism if $(a,b) \in R \iff (\phi(a), \phi(b)) \in S$ (in case $R$ and $S$ are binary relations). But then what are your homomorphisms in your example? Or do you only require $(a, b)\in R \implies (\phi(a), \phi(b)) \in S$? In that case $\phi(x) = \max{1, x-1}$ and $\psi(x)=x$ would work... But then what's stopping $\psi:B\to A$ from being an isomorphism? – Milten Mar 17 '21 at 22:51
  • @Milten I'm defining homomorphisms with a one-direction if, and isomorphisms as injective onto homomorphisms. – Navid Mar 18 '21 at 09:43
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    @Navid Ah, the right definition should be that the map and its inverse should both be homomorphisms (as in any category). As you wrote it, isomorphic-ness wouldn't be an equivalence relation. And with this your example works again – Milten Mar 18 '21 at 11:19
  • @Mitten I suppose this is the definition they use in universal algebra. I'm reading Dunn and Hardegree. – Navid Mar 19 '21 at 12:23
  • @Milten Looking more closely it seems to be a confusion in the book. – Navid Mar 19 '21 at 16:58

2 Answers2

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Consider the abelian groups $G = \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies) and $H = \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies).

There is a surjective homomorphism $G\to H$ which is the quotient map $\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ on the first component and the identity on all other components: $(a_1,a_2,a_3,\dots)\mapsto (\overline{a}_1,a_2,a_3,\dots)$ where $\overline{a}_1$ is the residue of $a_1$ mod $2$.

There is also a surjective homomorphism $H\to G$ which projects out the first component: $(a_1,a_2,a_3,\dots)\mapsto (a_2,a_3,a_4,\dots)$.

But these groups are not isomorphic. Note that the element $(1,0,0,\dots)$ in $H$ has order $2$ but is not divisible by $2$. On the other hand, every element of order $2$ in $G$ is divisible by $2$, since it has a $0$ or a $2$ in each component.

Alex Kruckman
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Let $G$ be an infinite group and let $H$ be a nontrivial proper subgroup. Let $A$ be the $G$-set that has countably infinitely many singleton orbits and countably infinitely many regular orbits (i.e., orbits isomorphic to ${}_GG$). Now let $B$ be constructed exactly like $A$, but include one more orbit isomorphic to ${}_G(G/H)$. $A$ is not isomorphic to $B$, since $B$ contains a point whose stabilizer is $H$ and $A$ does not. But there is a homomorphism of $A$ onto $B$ that collapses one regular orbit to ${}_G(G/H)$, and there is a homomorphism of $B$ onto $A$ that collapses the orbit ${}_G(G/H)$ onto a singleton.

Keith Kearnes
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  • I'll mention for other readers that this is very similar to and generalizes Alex Kruckman's example if we set $G=\mathbb Z/4\mathbb Z$ and $H=2G$. The main difference being that Alex Kruckman makes homomorphisms of groups whereas Keith Kearnes makes homomorphisms of $G$-sets (but I'm pretty sure the constructions are equivalent). Also Alex doesn't have the singleton orbits. – Milten Mar 17 '21 at 23:53
  • KeithKearnes; Is it necessary for $G$ to be infinite? – Milten Mar 17 '21 at 23:58
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    It is not necessary for $G$ to be infinite. I picked it that way to ensure that it had a nontrivial proper subgroup, but for this you only need that $|G|$ is not prime. – Keith Kearnes Mar 18 '21 at 00:13