The question is as follows
Show that if $(\tilde{X_1}, p_1)$ and $(\tilde{X_2}, p_2)$ are covering spaces of a space $X$ such that for $x_i \in \tilde{X_i}$ with $p_1(\tilde{x_1}) = p_2(\tilde{x_2})=: x_0$ and $p_{1*}(\pi_1(\tilde{X_1}, \tilde{x_1}))$ and $p_{2*}(\pi_1(\tilde{X_2}, \tilde{x_2}))$ conjugate in $\pi_1(X, x_0)$, then the covering spaces are isomorphic.
So far, I have the following:
Since the subgroups are conjugate, there is a loop $f$ in $X$ at $x_0$ such that $$p_{1*}(\pi_1(\tilde{X_1}, \tilde{x_1})) = [\overline{f}]p_{2*}(\pi_1(\tilde{X_2}, \tilde{x_2}))[f]$$
Since $f$ is path in $X$ at $x_0$, there is a lift $g: I \to \tilde{X_2}$ of $f$ such that $g(0) = \tilde{x_2}$. It is easy to verify that the inverse path $\overline{g}$ is a lift of $\overline{f}$.
What I want to show is that
$$[\overline{f}]p_{2*}(\pi_1(\tilde{X_2}, \tilde{x_2}))[f] = p_{2*}(\pi_1(\tilde{X_2}, g(1)))$$
Knowing that $\overline{f} = p_2 \circ \overline{g}$ and $f = p_2 \circ g$, it isn't too difficult to show that $[\overline{f}]p_{2*}(\pi_1(\tilde{X_2}, \tilde{x_2}))[f] \subset p_{2*}(\pi_1(\tilde{X_2}, g(1)))$. However, I'm struggling with the reverse inclusion. Any help would be greatly appreciated.
Best Answer
What you have shown is that for any loop $f$ at $x_0$ and any lift $g$ of $f$ one has $$[\overline{f}]p_{2*}(\pi_1(\tilde{X_2}, g(0)))[f] \subset p_{2*}(\pi_1(\tilde{X_2}, g(1))) .$$ You use this for $g(0) = \tilde{x_2}$. But it also shows that for the loop $\bar f$ which has $\bar g$ as a lift, we have $$[\bar{\bar f}]p_{2*}(\pi_1(\tilde{X_2}, \bar g(0)))[\bar f] \subset p_{2*}(\pi_1(\tilde{X_2}, \bar g(1)))$$ which is the same as $$p_{2*}(\pi_1(\tilde{X_2}, g(1))) \subset [\overline{f}]p_{2*}(\pi_1(\tilde{X_2}, g(0)))[f] .$$