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Let $F_2=\langle a, b\rangle$ be the non-abelian free group on two generators. Let $\phi_1, \phi_2: F_2\to F_2$ be two group homormophisms. It may be vague, but I still want to ask the following question.

Is there any ``geometric" methods to test whether $\phi_i$'s are conjugate, i.e. $\phi_1(g)=h\phi_2(g)h^{-1}, \forall g\in F_2$ for some $h\in F_2$?

Here, for ``geometric methods", I do not have a good example in mind, but it roughly means some methods in geometric group theory by studying the coarse properties of there maps, maybe by studying good paths in the cayley graph of $F_2$ under these maps...

Of course, to check conjugacy, we only need to test it at the two generators, but I think this as an algebraic method.

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I myself tend to think of testing conjugacy of two elements of $F_2$ as a geometric operation. Let $R_2$ denote the rose with two petals labelled $a,b$.

Interpret the two words $\phi_1(a),\phi_2(a)$ as two maps $S^1 \mapsto R_2$ where $R_2$ denotes the rose with two petals labelled $a,b$. Ordinarily one thinks of these two maps as base-pointed paths with domain $[0,1]$ mapping each endpoint $0,1$ to the base point. But now we are ignoring the base point and hence thinking of the domain as $S^1$.

Tighten each of the two maps (by homotopy) to obtain two local embeddings $f_1,f_2 : S^1 \to R_2$. The two elements $\phi_1(a),\phi_2(a)$ are conjugate in $F_2$ if and only if $f_1,f_2$ are the same up to cyclic permutation.

What I have described can be done completely algebraically: ignoring the base point and tightening by homotopy is the same as cyclic cancellation. Nonetheless, the geometric point of view is often helpful.

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  • $\begingroup$ Thank you! could you please provide a reference for your 2nd sentence. it seems you are saying that for $u, v, w, g, h\in F_2$, the solution to the equation $ugu^{-1}vhv^{-1}=wghw^{-1}$ is $u^{-1}v=g^nh^m$ for some $n, m \in \mathbb{Z}$. $\endgroup$
    – ougao
    Commented Sep 29, 2016 at 12:49
  • $\begingroup$ I take back what I said about testing the images of two generators being insufficient, and have rewritten my answer accordingly. I was mistakenly thinking about rank $n \ge 3$, where testing the images of a free basis of $n$ elements is not sufficient to determine conjugacy. $\endgroup$
    – Lee Mosher
    Commented Sep 29, 2016 at 13:12

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