Relation between free homotopy and in the same conjugacy class I had the following question when doing Hatcher's exercise 1.1.6.:
Question: two fundamental group $\pi_1(X,x_0)$ and $\pi_1(X,y_0)$ are isomorphic, induced by the path connecting $x_0$ and $y_0$. Does this ismorphism preserves the conjugacy classes? (assume $X$ is connected)
Hatcher's exercise could be viewed as a special case with $x_0=y_0$, which allows us to use other loops in $\pi_1(X,x_0)$ and the retraction trick. In the general cases with $x_0\neq y_0$, I'm curious whether this is still true. In some concrete examples, such as $X$ being the "figure of 8", or $X=SU(2)/\mathbb{Q}$ with $\mathbb{Q}$ being the quaternion group, I could construct loops that satisfy the statement.
In my attempt of a general proof, I got stuck at the following place.
I know how to prove that "two loops $f$ and $g$ (without common points) are freely homotopic iff there is a path $p$ connecting $x_0\in f$ and $y_0\in g$ such that $f$ is homotopic to $pgp^{-1}$". But I don't know how to show $pgp^{-1}$ and $g$ are in the same conjugacy class.
I'm not sure whether the statement is true or not. I'll be grateful if anyone can provide some hints or references.
 A: Yes.
First note that a group isomorphism $\phi : \pi_1(X,x_0) \to \pi_1(X,y_0)$ is induced by any path $u$ from $x_0$ to $y_0$ via $\phi(g) = [u^{-1}] * g * [u]$. Here $*$ denotes composition of path homotopy classes. In general $\phi$ depends on the choice of $u$ (unless $X$ is simply connected), thus one should not say that $\phi$ is induced by the path connecting $x_0$ and $y_0$.
Your question is whether $\phi$ preserves the conjugacy classes. An equivalent question is whether for two loops $\gamma_i$ based at $x_0$ which are freely homotopic also the loops $u^{-1} * \gamma_i * u$ are freely homotopic. We should be aware that free homotopy of loops needs interpretation. We can regard loops $\gamma$ as pointed maps $(S^1,*) \to (X,x_0)$; then $\pi_1(X,x_0)$ consists of pointed homotopy classes of such maps. A free homotopy is then a homotopy which is not required to be basepoint-preserving.  We can also regard loops $\gamma$ as closed paths $I  \to X$ such that $\gamma(0)= \gamma(1) = x_0$; then $\pi_1(X,x_0)$ consists of their path homotopy classes. A free homotopy of closed paths (or a loop homotopy) is not required to be basepoint-preserving, but is assumed to produce a closed path at each time $t$. See the answer to Characterizing simply connected spaces for a discussion.
In the "path interpretation" it is very easy to see that each loop $\gamma$ based at $x_0$ is freely homotopic to the loop $u^{-1} * \gamma * u$ based at $y_0$ which answers your question. Explicitly we define
$$H :  I \times I \to X, H(s,t) = \begin{cases} u(t(1 - 3s)) & s\le 1/3 \\ \gamma(3s - 1) & 1/3 \le s \le 2/3 \\ u(t(3s-2)) & s \ge 2/3 \end{cases}.$$
This is a loop homotopy from $c * \gamma *c$ to $u^{-1} * \gamma * u$, where $c$ is the constant path at $x_0$. But clearly $c * \gamma *c$ and $\gamma$ are path homotopic, in particular loop homotopic.
Update:
That $\phi$ preserves conjugacy classes can easily  be seen directly (without using free homotopies). In fact, each group homomorphism $\psi : A \to B$ preserves conjugacy classes:
Let $a_1, a_2 \in A$ be conjugate in $A$, i.e. $a_2 = g^{-1}a_1g$ for some $g \in A$. Then $\psi(a_2) = \psi(g)^{-1}\psi(a_1)\psi(g)$. Hence $\psi(a_1), \psi(a_2) \in B$ are conjugate in $B$.
Now recall that $\phi$ is a group isomorphism. Therefore $\phi$ induces a bijection between conjugacy classes in $\pi_1(X,x_0)$ and conjugacy classes in $\pi_1(X,y_0)$.
However, I think it is an additional interesting information that each loop $\gamma$ based at $x_0$ is freely homotopic to the loop $u^{-1} * \gamma * u$ based at $y_0$.
