I will start with a general algorithmic question:

**Question.** (A faithfulness decision problem.) Suppose that $G, H$ are finitely-presented groups with decidable word problem. Is injectivity decidable for homomorphisms $f: G\to H$ (defined by the images of generators of $G$)?

[A very minor side issue: I am not even sure if this decision problem has an established name. For some reason, Max Dehn did not consider it.]

The non-injectivity problem is trivially semidecidable.

More specifically, I am interested in decidability of faithfulness for homomorphisms from (finitely generated, nonabelian) free groups to (nonelementary) hyperbolic groups $H$. The decision problem then becomes:

**Freedom problem.** Do given elements $g_1,...,g_n\in H$ generate a rank $n\ge 2$ free subgroup?

Things that I know:

If $H$ is

locally quasiconvex(or, more generally, all finite rank free subgroups of $H$ are undistorted), then the freedom problem is decidable. The same applies to the injectivity problem for homomorphisms from general finitely-presented groups $G\to H$ to locally quasiconvex hyperbolic groups. The injectivity problem is decidable if $H$ is the fundamental group of a compact hyperbolic 3-manifold.

It is natural to expect that the decidability of the freedom problem in hyperbolic groups $H$ is closely related to distortion of free subgroups in $H$. One can conjecture that if every f.g. free subgroup in $H$ is recursively distorted then the freedom problem is decidable. One can also conjecture that hardness of the freedom problem correlates to the degree of distortion of free subgroups. (For instance, for "hyperbolic hydra groups" of Brady-Dison-Riley, the distortions of some free subgroups are given by Ackerman functions.) I do not know how to approach any of these conjectures.

So, my question is if anything else is known about these problems.

One more thing: There is a "dual" problem about **surjectivity** of group homomorphisms. I am not asking about this one. (It is more related to the subgroup membership problem and more is known about it.)