This is about exercise V.4.1. in Massey's A Basic Course in Algebraic Topology. Section V.4. is about the fundamental group of a covering space. Massey assumes that all spaces involved are both arc-connected and locally arc-connected.
Theorem 4.1. states that for a covering space $p: \tilde{X} \to X$, $\tilde{x}_0 \in \tilde{X}$ and $x_0 = p(\tilde{x}_0)$ the induced morphism $p_*:\pi(\tilde{X},\tilde{x}_0) \to \pi(X,x_0)$ is a monomorphism. What follows is a discussion on what happens if one replaces $\tilde{x}_0$ by another $\tilde{x}_1$ which is also mapped to $x_0$ via $p$. The result is that $p_*(\pi(\tilde{X},\tilde{x}_0))$ and $p_*(\pi(\tilde{X},\tilde{x}_1))$ are conjugate subgroups of $\pi(X,x_0)$ and that all subgroups of $\pi(X,x_0)$ which lie in the conjugacy class of $p_*(\pi(\tilde{X},\tilde{x}_0))$ arise this way. This is the content of
Theorem 4.2. Let $p:\tilde{X} \to X$ be a covering space and $x_0 \in X$. Then the subgroups $p_*(\pi(\tilde{X},\tilde{x}))$ for $\tilde{x} \in p^{-1}(x)$ are exactly a conjugacy class of subgroups of $\pi(X,x)$.
I'm having problems with
Exercise 4.1 Discuss the effect of changing the "base point" $x_0$ in the statement of Theorem 4.2. to a new base point $x_1 \in X$.
My first idea would be to pick some path connecting $x_0$ and $x_1$ in $X$ (it exists by the assumption of path-connectedness). This would yield an isomorphism of groups $\pi(X,x_0) \to \pi(X,x_1)$. Under this isomorphism the conjugacy class coming from Theorem 4.2. for $x_0$ maps to a conjugacy class of subgroups in $\pi(X,x_1)$. Is the content of the exercise to compare this class to the class of subgroups of $\pi(X,x_1)$ that I get when I use Theorem 4.2. with $x_1$ in place of $x_0$? If so, could anybody provide a hint please?
edit: Here is the diagram I have been talking about in the comments:
$\pi(\tilde{X},\tilde{x}_0) \longrightarrow \pi(\tilde{X},\tilde{x}_1)$
$\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$
$\pi(X,x_0) \longrightarrow\,\, \pi(X,x_1)$
The vertical maps are $p_*$, the lower horizontal one is conjucation with the class of a fixed path from $x_0$ to $x_1$ in $X$ and the upper horizontal map is conjugation with the class of the lift of said path.
Is there something else to this exercise? If I now change $\tilde{x}_0$ in the fiber $p^{-1}(x_0)$ and I then change the lifting of the path from $x_0$ to $x_1$ to $\tilde{X}$ accordingly (i.e. changing its starting point), then I run through a list of commutative squares as above, one for each $\tilde{x_0} \in p^{-1}(x_0)$. Doing so I run through the whole conjugacy class of $p_*\pi(\tilde{X},\tilde{x_0})$ in $\pi(X,x_0)$ in the left half of the diagram. Is the question now what happens on the right half of the diagram when I do so? My guess is that I simply run through the whole conjugacy class of $p_*\pi(\tilde{X},\tilde{x_1})$ in $\pi(X,x_1)$. Is that correct?
Did I make myself clear? I was fixing $\tilde{x_0}$ first and then left it run again, hopefully I did not lose you with that.