Is this surjective homomorphism necessarily injective? (Free Product of Cyclic groups)

Question

Let $G=\langle g_1,\dots,g_m\mid g_1^{k_1}=\dots=g_m^{k_m}=1\rangle$. (Where $k_1,\dots,k_m$ are integers.)

Let $\psi:G\to G$ be a surjective homomorphism.

Is $\psi$ necessarily injective?

In other words, if $x\in\ker\psi$, then is $x=1$?

Thanks.

"Intuitively" it seems true that $\psi$ is injective, but I can't seem to find a proof for it.

Answer 1

Yes. In this setting, $\psi$ is injective if it is also surjective. That is, free products of finitely many cyclic groups (in fact, free products of finitely many finite groups) are Hopfian.*

One reason for this is the following two results:

Lemma 1. Free products of finitely many finite groups are virtually free, and hence residually finite.

Lemma 2. Finitely generated, residually finite groups are Hopfian.

I once wrote out the proof of Lemma 2 in this old answer.

You can find a direct proof that your groups are Hopfian, which does not apply Lemma 2, on MathOverflow.

*Examples of finitely presented, non-Hopifan groups can be found here.