The question arises from the proof of Galois covering theorem, the one describing the one-one correspondence of intermediate coverings (equivalent class) and subgroups of the deck transformation group of a certain Galois covering. I type the theorem below first.
Fix a Galois covering $p:Y\to X$, where we assume $Y$ is connected while $X$ is locally connected, and denote the deck transformation group of a covering by ${\rm Deck}(Y/X)$. And define the intermediate covering $q:Z\to X$, $f:Y\to Z$, s.t. diagram
commutes, where $Z$ is connected space with $f$ and $q$ both covering maps. Two such diagrams ($p:Y\to X$ is fixed, different $Z,f,q$ and $Z^{\prime},f^{\prime},q^{\prime}$) are said to be equivalent if there exists a homeomorphism $\varphi:Z\to Z^{\prime}$, s.t. the diagram
commutes. Now we have two sets, the equivalent class of intermediate covering maps $\mathcal{C}=\{\text{equivalent diagrams of intermediate coverings of a fixed Galois covering }p:Y\to X\}$ and all the subgroups of ${\rm Deck}(Y/X)$: $\mathcal{G}=\{H|H<{\rm Deck}(Y/X)\}$, we tentatively define 2 maps between these groups, ${\rm Deck}:\mathcal{C}\to \mathcal{G}$ by:
One can check this is well-defined. To define the second map, we note first the natural action of ${\rm Deck}(Y/X)$ on $Y$ by the definition of homeomorphism $g(y),\ g\in {\rm Deck}(Y/X)$, so is one of its subgroup $H$. Construct the quotient group and canonical map $\pi:Y\to Y/H$,and by the universal property of quotient, there exits unique map $\bar{p}:Y/H\to X$, s.t.
commutes. It can be proved that $\pi$ and $\bar{p}$ are also covering maps, which is a part of the theorem below. Now define the second map $/:\mathcal{G}\to \mathcal{C}$, by:
To sum up, we define two maps $\mathcal{C}\underset{/}{\stackrel{\rm Deck}{\rightleftharpoons}}\mathcal{G}$. Now states the theorem:
Fix a Galois covering $p:Y\to X$, where $Y$ is connected and $X$ is locally connected. Then (1) any covering $q:Z\to X$, Z connected, s.t. there exists morphism $f:Y\to Z$ to commutes the diagram
then the map $f$ is a Galois covering. (2) ${\rm Deck}\circ /={\rm id}_{\mathcal{G}}$ and $/\circ {\rm Deck}={\rm id}_{\mathcal{C}}$ (3) under Deck and / defined above, covering $q:Z\to X$ is Galois iff ${\rm Deck}(Y/Z)\vartriangleleft{\rm Deck}(Y/X)$.
Having proved the first two parts, we can replace the general intermediate covering $q:Z\to X$ and $f:Y\to Z$ by $\bar{p}:Y/H\to X, \text{for a certain H}\in \mathcal{G}$ and $\pi:Y\to Y/H$, because of the 1-1 correspondence in (2). Then the theorem's third part asks equivalently when is $\bar{p}$ also Galois? Iff $H\vartriangleleft {\rm Deck}(Y/X).$ The proof is to figure out the relation between ${\rm Deck}(Y/X)$ and ${\rm Deck}\big((Y/H)\big/X\big)$. When$H\vartriangleleft {\rm Deck}(Y/X)$ we can induced a natural homeomorphism from $g\in {\rm Deck}(Y/X)$ on $Y/H$, called $\bar{g}$, by $\bar{g}(\bar{y}):=\overline{g(y)},\ any\ y\in Y$. One can check that the normality guarantees $\bar{g}$ is well defined and the definition is to say $\bar{g}\circ\pi=\pi\circ g$, which makes the diagram
commute. This definition further induces a homomorphism from ${\rm Deck}(Y/X)$ to ${\rm Deck}\big((Y/H)\big/X\big)$ Then here arises the question, if this homomorphism is onto, we will finish the proof of necessity part of the theorem (I omit the proving detail). And more or less we can guess ${\rm Deck}(Y/X)\big/H={\rm Deck}\big((Y/H)\big/X\big)$ from fundamental theorem of group homomorphism. So the question becomes for any $h\in {\rm Deck}\big((Y/H)\big/X\big)$, is there $g\in {\rm Deck}(Y/X)$, s.t. $h\circ \pi=\pi\circ g$ to let the diagram
commutes?
At first I tried to construct $g$ from the property of quotient but failed, then I consider $\pi$ is a covering. For any $\bar{y}\in Y/H$, we can choose an open neighborhood $U$ of $\bar{y}$ s.t. $\pi^{-1}(U)=\coprod_{\alpha\in I}V_{\alpha}$ where $V_{\alpha}$ is open in $Y$ and $\pi\Big|_{V_{\alpha}}$ is homeomorphism, with neighborhood $hU$ of $h(\bar{y})$ s.t. $\pi^{-1}(hU)=\coprod_{\beta\in I}W_{\beta}$ where $W_{\beta}$ is open in $Y$ and $\pi\Big|_{W_{\beta}}$ is homeomorphism. Then I consider lift $h$ on those disjoint union piece by piece $g\Big|_{V_{\alpha}}:V_{\alpha}\to W_{\beta},\ g=\Big(\pi\Big|_{V_{\alpha}}\Big)^{-1}\circ h\circ\pi\Big|_{W_{\beta}}$, and for two such open sets of $Y/H$ that intersect $U\cap U_1\neq \emptyset$, choose the lift to match on those disjoint pieces and by gluing lemma, I get a larger lifting $g\Big|_{V_{\alpha}\cup V_{1\alpha}}:V_{\alpha}\cup V_{1\alpha}\to W_{\beta}\cup W_{1\beta}$. So after I repeat the same process on an open cover of $Y/H$, I somehow "glue" a $g$ on $Y$ which satisfies the requirement. But that means I may use gluing lemma infinite times, which may not be right then.
Then I get stuck. Any hint or help? Thanks a lot.