# Subalgebras isomorphic to their images

Question: Given a homomorphism $$f:A \to B$$, where $$A$$ and $$B$$ are algebras in any variety (in the sense of universal algebra), is it true that if $$f(S)$$ is isomorphic to $$S$$ for all subalgebras $$S$$ of $$A$$ (in symbols: $$\forall S \le A \ f(S) \cong S$$), then $$f$$ is necessarily injective?

For some specific varieties, the answer can easily be shown to be "yes":

• Sets: Consider the 2-element subsets of $$A$$.
• Groups (or modules or vector spaces or rngs (possibly non-unital rings)): Consider the kernel of $$f$$.
• Lattices (not necessarily with top or bottom elements): For any $$a_1, a_2 \in A$$, consider the sublattice $$\{a_1, a_2, a_1 \land a_2, a_1 \lor a_2\}$$ of $$A$$.

For $$n\in\mathbb{Z}$$, let $$[n]=\{x\in\mathbb{Z}:x\geq n\}$$ and let $$A$$ be the set of all functions $$[n]\to\{0,1\}$$ which are eventually $$0$$, where $$n$$ can be any integer. Let $$f:A\to A$$ be the map that restricts a function $$[n]\to\{0,1\}$$ to $$[n+1]$$. Let $$g:A\to A$$ be given by $$g(a)(x)=a(x+1)$$ (so $$g$$ maps functions $$[n]\to\{0,1\}$$ to functions $$[n-1]\to\{0,1\}$$) and let $$h:A\to A$$ be given by $$h(a)(x)=a(x)$$ if $$x\neq 0$$ and $$h(a)(0)=1-a(0)$$ (so $$h$$ maps functions $$[n]\to\{0,1\}$$ to functions $$[n]\to\{0,1\}$$). Note that $$g$$ is a bijection, and that $$g,g^{-1},$$ and $$h$$ all commute with $$f$$.
Now let us consider $$A$$ as an algebra with respect to the unary operations $$g,g^{-1}$$, and $$h$$. Then $$f:A\to A$$ is a homomorphism which is surjective but not injective. I claim that the only nonempty subalgebra of $$A$$ is $$A$$ itself, and so $$f:A\to A$$ is a counterexample to your question.
To prove this, just observe that any element of $$A$$ can be taken to any other element by repeated application of $$g,g^{-1}$$, and $$h$$. We can first use $$g$$ or $$g^{-1}$$ to assume the two elements have the same domain, and then we can use $$h$$ conjugated by $$g$$ any number of times to modify the finitely many values on which the functions differ. So, any element of $$A$$ generates the whole algebra.
Here is another perspective on this example. You could instead consider $$A$$ as a $$G$$-set where $$G$$ is the group of permutations of $$A$$ generated by $$g$$ and $$h$$. This group can be identified with the lamplighter group in an obvious way, and then $$A$$ can be identified with the coset space $$G/H$$ where $$H$$ is the subgroup generated by the elements $$g^nhg^{-n}$$ for all $$n>0$$. The map $$f$$ then sends a coset $$xH$$ to $$xg^{-1}H$$. This is well-defined since $$gHg^{-1}\subseteq H$$ but not injective since $$gHg^{-1}$$ is a proper subset of $$H$$.
In the same way, then, you could let $$G$$ be any group with a subgroup $$H$$ and an element $$g\in G$$ such that $$gHg^{-1}$$ is strictly contained in $$H$$. Then taking $$A$$ to be the $$G$$-set $$G/H$$, the map $$f:A\to A$$ given by $$f(xH)=xg^{-1}H$$ is a surjective but not injective homomorphism, but $$A$$ has no nonempty proper subalgebras.