# Bijective homomorphism which is not an isomorphism

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.)

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 $$0\in \mathbb{N}$$.
• So the signature contains one relation symbol, namely whether an element belongs to $\mathbb N$? Nov 13, 2021 at 12:42
• @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? Nov 13, 2021 at 12:44