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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.

Edward
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    I know what it means for two elements of the same group to be conjugate. But what does it means for two elements $f \in \pi_1(X,x_0)$ and $g \in\pi_1(X,y_0)$ in two different groups to be conjugate? – Lee Mosher Jun 13 '21 at 21:48
  • @ Lee Sorry for the confusion. Maybe I can ask my question in this way directly: $\pi_1(X,x_0)$ and $\pi_1(Y,y_0)$ are isomorphic to each other, induced by the paths connecting $x_0$ and $y_0$, whether this isomorphism preserves conjugacy class. Does this sound good to you? – Edward Jun 13 '21 at 22:46
  • Then you should edit your question to make precise what you really want to know. – Paul Frost Jun 13 '21 at 22:57
  • @ Paul Thanks for reminding. I have edited my question. – Edward Jun 13 '21 at 23:10

1 Answers1

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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 $y_0$.

Paul Frost
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  • Thanks a lot for your answer! Maybe one naive question: would you mind explaining why the free homotopy between $\gamma$ and $u^{-1}\star\gamma\star u$ implies that they are in the same conjugacy class? Namely, why the conjugacy class is a invariant during this smooth interpolation $H$ you constructed. – Edward Jun 16 '21 at 14:16
  • Let $\gamma_i$ be loops based at $x_0$. Then you know that $[\gamma_i]$ are conjugate in $\pi_1(X,x_0)$ iff the $\gamma_i$ are freely homotopic. Similarly $[u^{-1} * \gamma_i * u]$ are conjugate in $\pi_1(X,y_0)$ iff the $u^{-1} * \gamma_i * u$ are freely homotopic. Now just use the fact that $\gamma_i $ and $u^{-1} * \gamma_i * u$ are freely homotopic. – Paul Frost Jun 16 '21 at 17:01
  • Thanks! Sorry for being a bit repetitive. I understand your (first) statement. For your second statement "Now just use the fact....", how to show that free homotopy between $\gamma_i$ and $u^{-1}\star \gamma_i \star u$ requires $[\gamma_i]$ and $[u^{-1}\star \gamma_i \star u]$ to be in the same class? I can only construct examples that $[\gamma_i]$ and $[u^{-1}\star \gamma_i \star u]$ are different group elements within the same class, but wondering whether it is possible that they belong to two different classes with the same number of group elements? – Edward Jun 16 '21 at 19:20
  • Assume that (1) $[\gamma_i]$ are conjugate in $\pi_1(X,x_0)$. We want to show that (2) $[u^{-1}\gamma_i u]$ are conjugate in $\pi_1(X,y_0)$. (1) means that the $\gamma_1 \simeq_{free} \gamma_2$. But $\gamma_i \simeq_{free} u^{-1}* \gamma_i * u$. Hence $u^{-1}* \gamma_1 * u \simeq_{free} u^{-1}* \gamma_2 * u$. This proves (2). – Paul Frost Jun 16 '21 at 20:38
  • Thanks! I totally agree with your derivation. Let me be more concrete. Suppose $\pi_1(X)$ has three classes: the identity $\mathbb{I}$ and two other classes $A$ and $B$. $A$ and $B$ has the same number of elements. Is it possible for $[\gamma]$ to belong to class $A$ while $[u^{-1}\star \gamma \star u]$ belong to the class $B$? – Edward Jun 16 '21 at 22:00
  • @Edward Which classes do you mean? Conjugacy classes? If so, then $\phi : \pi_1(X,x_0) \to \pi_1(X,y_0)$ induces a bijection of conjugcy classes as we have seen, but nevertheless a conjugacy class $A(x_0)$ in $\pi_1(X,x_0)$ is not the same as a conjugacy class $A(y_0)$ in $\pi_1(X,y_0)$. Thus it does make sense to ask whether $[\gamma] \in A(x_0)$ but $[u^{-1} \gamma u] \notin A(x_0)$. – Paul Frost Jun 17 '21 at 10:42