A simple undecidable problem for free groups (Q6549687)

Let \(A^{\ast}\) and \(B^{\ast}\) be two free monoids, \(h,g: A^{\ast} \rightarrow B^{\ast}\) be two morphisms and let \(\mathrm{Eq}(g,h)=\{ w \mid g(w)=h(w) \}\) be theirs equalizer. Whether or not \(\mathrm{Eq}(g,h)\) is trivial is an undecidable problem (the Post correspondence problem, see [\textit{E. L. Post}, Bull. Am. Math. Soc. 52, 264-268 (1946; Zbl 0063.06329)]). For free groups, the related problems are more complicated due to cancellations by the inverse elements.\N\NLet \(F_{n}\) be the free group on \(n\) generators. In the paper under review, the author shows that is undecidable for two morphisms \(g,h : F_{n}\rightarrow F_{2}\) and a generator element \(a\) of \(F_{2}\), whether or not there exists an element \(w \in F_{n}\) such that \(g(w)=a\) and \(h(w)=1\).