---- > [!theorem] Theorem. ([[characterization of covering space equivalence (matched base point)]]) > Let $p: E \to B$ and $p':E \to B$ be [[covering space|covering maps]]; let $p(e_{0})=p'(e_{0}')=b_{0}$. There is an [[isomorphism of covering spaces|equivalence]] $h: E \to E'$ such that $h(e_{0})=e_{0}'$ [^1] if and only if the [[group|groups]] $H_{0}=p_{*}'\big( \pi_{1}(E', e_{0}) \big) \text{ and } H_{0}'=p_{*}\big(\pi_{1}(E,E_{0}')\big)$ > are equal. If $h$ exists, it is unique. > [!proof]- Proof. ([[characterization of covering space equivalence (matched base point)]]) > Let's first prove the "only if". Given $h$, $p_{*}\big( \pi_{1}(E,e_{0}) \big)=p_{*}'h_{*}\big( \pi_{1}(E,E_{0}) \big)$; [[homeomorphic topological spaces have isomorphic fundamental groups|since]] $h$ is a [[homeomorphism]] we have $h_{*}\big( \pi_{1}(E,E_{0}) \big)=\pi_{1}(E',e_{0}')$. Thus $H_{0}=H_{0}'$. > > Now prove the "if" part. This follows from appplying [[the general lifting lemma for covering spaces|the general lifting lemma]] (four times). Assume $H_{0}=H_{0}'$, WTS $h$ exists. Start here: > > ![[CleanShot 2024-04-20 at 09.51.10.jpg]] > > $p'$ is a [[covering space|covering map]] and $E$ is [[path-connected]] and [[locally connected, locally path-connected|locally path-connected]], and $H_{0} \subset H_{0}'$, there exists a unique map $h:E \to E'$ with $h(e_{0})=e_{0}'$ that is a lifting of $p$ (that is, such that $p' \circ h = p$). > > Reversing roles of $E,E'$ (since $H_{0}' \subset H$) lets us likewise obtain $k:E' \to E$ with $k(e_{0}')=e_{0}$ such that $p \circ k=p'$. Now consider > ![[CleanShot 2024-04-20 at [email protected]]] > The map $k\circ h$ is a lifting of $p$ (since $p \circ k \circ h=p' \circ h=p$) with $p(e_{0})=e_{0}$. The [[identity map]] $\id_{E}$ of $E$ is another such lifting. By the uniqueness part of [[the general lifting lemma for covering spaces]], we must have $k \circ h=\id_{E}$. A similar argument will show that $h \circ k=\id_{E'}$. So $h$ is a [[homeomorphism]] with inverse $k$, and $h$ is unique. ---- #### [^1]: This is a subtle condition. Munkres relaxes it later (in the general case, we need/get that the [[subgroup|subgroups]] are [[conjugate]]). ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```