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.
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 $y_0$.