Bijective homomorphism which is not an isomorphism

Question

I have the following question:

Find an example of a bijective homomorphism between two structures $\underline{A}$ and $\underline{B}$ which is not an isomorphism.

I answered this by taking the identity map between an algebra and a relational structure with same domain.

However, I want to find an example when $\underline{A}=\underline{B}$.

I could not come up with one, neither could I prove that no such example exists. Any hints?

(All definitions are from https://math.stackexchange.com/a/2170754/266110.)

score 8 · Accepted Answer · answered Nov 13 '21 at 12:40

8

Consider a first-order language $L$ with a single unary predicate symbol $P$, and make the set $A=\mathbb{Z}$ into an $L$-structure $\underline{A}$ by taking $P^{\underline{A}}=\mathbb{N}$. Then consider the map $f:A\to A$ given by $a\mapsto a+1$; this is a bijective homomorphism, since $a\in \mathbb{N}\implies f(a)\in \mathbb{N}$ for each $a\in A$. But it is not an isomorphism, since $f^{-1}(0)=-1\notin \mathbb{N}$ even though $0\in \mathbb{N}$.

answered Nov 13 '21 at 12:40

Atticus Stonestrom

11,426

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So the signature contains one relation symbol, namely whether an element belongs to $\mathbb N$? – modeltheory Nov 13 '21 at 12:42
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@modeltheory precisely! :) a unary predicate in a first-order structure is really just the same thing as a distinguished subset; in this case, we are declaring that $\underline{A}\models P(a)$ iff $a\in\mathbb{N}$. does that make sense? – Atticus Stonestrom Nov 13 '21 at 12:44
Yes, that makes a lot of sense. Thank you! – modeltheory Nov 13 '21 at 12:44
@modeltheory my pleasure, happy it helped! – Atticus Stonestrom Nov 13 '21 at 12:45

Bijective homomorphism which is not an isomorphism

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