Let $p: E\rightarrow B$ and $p':E' \rightarrow B$ covering map, E and E' path-connected and locally path-connected, isomorphismus between E and E'

Question

Let $p: E\rightarrow B $ with $b_0 \in B, e_0\in E$ such that $p(e_0)=b_0$ a covering space. Let $p':E' \rightarrow B$ covering maps, and let $e'_0 \in E' $ such that $p'(e'_o)=b_0$. Let now E and E' path-connected and locally path-connected. Show that $E \cong E' \Leftrightarrow p'_*(\pi_1(E',e'_0))$ and $p_*(\pi_1(E,e_0))$ are conjugte in $\pi(B,b_0).$

I want to use the following theorem but I'm not sure if all the preconditions are fullfilled.

$\textbf{Theorem}$: Let $p: E \rightarrow B$ and $p': E' \rightarrow B $ two covering spacaes, with $b_0 \in B, e_0\in E, e'_0 \in E'$ such that $p(e_0)=b_0=p'(e'_0)$ and let E, E' path-connected and locally path-connected. Then \begin{align*} \exists \, \, \phi:E \rightarrow E' \, \, \text{isomorphismus with } \, \phi (e_0)=e'_0 \Longleftrightarrow p_*(\pi_1(E,e_0))=p'_*(\pi_1(E',e'_0)). \end{align*}

Any hint or suggestions will be really apprecieted.

Answer 1

You have to use also the following two statements:

$\textbf{Lemma 1}$: If $\tilde{e_0} \in p^{-1}(b_0)$, then $p_{*}(\pi_1(E, e_0))$ is conjugate to $p_{*}(\pi_1(E, \tilde{e_0}))$ in $\pi_1(B, b_0))$.

$\textbf{Lemma 2}$: Let $H$ a subgroup of $\pi_1(B, b_0)$ conjugate to $p_{*}(\pi_1(E, e_0))$. Then it exists a point $\tilde{e_0} \in p^{-1}(b_0)$ such that $H = p_{*}(\pi_1(E, \tilde{e_0}))$.

"$\Rightarrow$": Since $E, E'$ are isomorphic, then it exists an isomorphismus $\phi: E \rightarrow E'$. So, by surjectivity of $\phi$ it exists an $\tilde{e_0} \in E$ s.t. $\phi(\tilde{e_0}) = e_0 '$. Then, by your theorem, you obtain that $p_{*}(\pi_1(E, \tilde{e_0})) = p'_{*}(\pi_1(E', e_0 '))$. Moreover, since $\phi$ is an isomorphism, then $p'(\phi(t)) = p(t) \, \, \forall t \in E$. For $t= \tilde{e_0}$ you have that $p'(\phi(\tilde{e_0}))= p'(e_0') = b_0$, so $b_0 \in p^{-1}(b_0)$. Then by Lemma 1, you obtain that $p_{*}(\pi_1(E, e_0))$ is conjugate to $p_{*}(\pi_1(E, \tilde{e_0}))$ in $\pi_1(B, b_0))$, hence the result.

"$\Leftarrow$": Since $p'_{*}(\pi_1(E', e_0'))$ is conjugate to $p_{*}(\pi_1(E, e_0))$ in $\pi_1(B, b_0)$, then by Lemma 2, it exists $\tilde{e_0} \in p^{-1}(b_0)$ s.t. $p'_{*}(\pi_1(E', e_0' ))= p_{*}(\pi_1(E, \tilde{e_0}))$. So by your theorem, it exists an isomorphism $\phi: E \rightarrow E'$ s.t. $\phi(\tilde{e_0})= e_0'$, hence the result.

Answer 2

Hints: If those subgroups are conjugate, then up to some path in $E'$ (and hence changing the base point), they are equal. You can then use the theorem.

If there is such an isomorphism, then up to a change of base point the subgroups are equal, hence they are conjugate if you fix the base points.