Is the inverse image of a group also a group for semigroup homomorphisms

Question

If $\varphi : S \to T$ is a surjective semigroup homomorphism between semigroups and $G \subseteq T$ is a group, then is $\varphi^{-1}(G)$ also a group?

I know that this result holds if $S$ and $T$ are finite, as then I can find an idempotent element in $\varphi^{-1}(1_G)$ and everything follows from that. But I cannot prove it for infinite semigroups, nor can I find a counter-example.

Answer 1

Project $\Bbb{N}$ on $\{0,1\}=\Bbb{Z}/(2)$, sending evens to $0$ and odds to $1$. So, no, the inverse image of a group under a semigroup homomorphism need not be a group, even if it is a surjective monoid homomorphism.

Answer 2

Your statement

I know that this result holds if $S$ and $T$ are finite, as then I can find an idempotent element in $\varphi^{-1}(1_G)$ and everything follows from that.

is unfortunately wrong. Let $T$ be the trivial group $1$, let $S$ be any finite non-group semigroup and let $\varphi:S \to T$ be the trivial map. Then $\varphi^{-1}(1) = S$ and thus $\varphi^{-1}(1)$ is not a group.

The correct statement is

Let $S$ be finite semigroup and let $\varphi : S \to T$ be a surjective semigroup morphism. Then for every group $G \subseteq T$, there exists a group $H \subseteq S$ such that $\varphi(H) = G$.