Is my alternative construction of a covering space $p:Y\to X$ with $p^\ast(\pi_1(Y))=H$, for an arbitrary $H\le\pi_1(X)$, correct? $\newcommand{\deck}{\operatorname{Deck}}$In Hatcher's Algebraic Topology, for a covering space $p:(\widetilde{X},\widetilde{x_0})\to(X,x_0)$ the image subgroup $H:=p^\ast(\pi_1(\widetilde{X},\widetilde{x_0}))\le\pi_1(X,x_0)$ is often discussed.
When $X$ is globally and locally path connected, and semi-locally simply connected, Hatcher shows that:

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*There is a path connected covering space $p_H:(X_H,x_H)\to(X,x_0)$ with $p_H^\ast(\pi_1(X_H,x_H))=H$ for any $H\le\pi_1(X,x_0)$

*Any two path connected covering spaces $q_1:(Y_1,y_1)\to(X,x_0),\,q_2:(Y_2,y_2)\to(X,x_0)$ with $q_1^\ast(\pi_1(Y_1,y_1))=q_2^\ast(\pi_1(Y_2,y_2))$ are isomorphic

The proof goes by taking a certain quotient of the universal covering space $\upsilon:(U,u_0)\to(X,x_0)$. He also discusses how to take quotients of $U\times\upsilon^{-1}(x_0)$ to identify all covering spaces of $X$ up to isomorphism, and how to take quotients $Y\twoheadrightarrow Y/G$ to find covering spaces with specified deck transformation groups, where $G$ has a covering space action on $Y$.
I noted that we could bring some of these ideas together to get a 'new' way of generating covering spaces with arbitrary image subgroups $H$. By the above classification theorem, this is not really new and will give you the same thing as $X_H$. However, the abstract formulation of $X_H$ is a little unwieldy, so I would hope that I've found a nice alternative description.
My question is, is my alternative construction correct?

Assumptions: $(X,x_0)$ is a nonempty, locally and globally path connected space with a simply connected covering space $\upsilon:(U,u_0)\to(X,x_0)$.
We know that $\pi_1(X,x_0)\cong\deck(U)$ via the following association:

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*For any loop $\lambda$ based at $x_0$, the unique lift of lambda to a path $\widetilde{\lambda}$ in $U$ that begins at $u_0$ must end at some point $\widetilde{\lambda}(1)\in\upsilon^{-1}(x_0)$; there is a unique deck transformation $\tau$ that takes $x_0$ to $\widetilde{\lambda}(1)$, and we map $[\lambda]\mapsto\tau$.

This defines a group isomorphism. Suppose $H\le\pi_1(X,x_0)$ is given. This group isomorphism identifies $H$ with a subgroup $\widetilde{H}\le\deck(U)$.
Since $U$ is path connected, any subgroup of the deck group has a covering space action over $U$: then, the quotient: $$q:U\twoheadrightarrow U/\widetilde{H}$$Is a covering map, and the associated deck group is $\widetilde{H}$ and $\pi_1(U/\widetilde{H},q(u_0))\cong\widetilde{H}$ by the isomorphism I mention above. We can see $\upsilon$ induces a map $p:(U/\widetilde{H},u_0)\to(X,x_0)$. I claim this is a covering map.
For, take $x\in X$ and any $\upsilon$-evenly covered neighbourhood $V$ of $x$. With no loss of generality, $V$ may be taken to be (path) connected. We have $p^{-1}(V)=qq^{-1}p^{-1}(V)=q(\upsilon^{-1}(V))$ by surjectivity. For any $h\in\widetilde{H}$ and any sheet $\widetilde{V}\subseteq\upsilon^{-1}(V)$, we can see $h(\widetilde{V})$ must also be a sheet for connectivity reasons. It then follows that if $a\in V_1$ is identified with $b\in V_2$ - for two sheets $V_{1,2}$ - then all of $V_1$ is identified with $V_2$, over $V$, by the same $h\in\widetilde{H}$. It's tedious to prove it all formally, but from this it's obvious that $q(\upsilon^{-1}(V))$ is a disjoint union of open sets all homeomorphic to $V$ via $p$, so $p$ is a covering.
A loop $\gamma$ in $U/\widetilde{H}$, based at $q(u_0)$, has a unique lift to a path $\widetilde{\gamma}$ in $U$ that begins at $u_0$, and there is a unique $h\in\widetilde{H}$ that takes $u_0$ to $\widetilde{\gamma}(1)$, which is independent of the representative $\gamma$ for the based homotopy class of $[\gamma]$. I can write: $$\pi_1(U/\widetilde{H},q(u_0))=\{[q\lambda]:\lambda\text{ is a path in $U$ from $u_0$ to $h(u_0)$ for some $h\in\widetilde{H}$}\}$$Then: $$p^\ast(\pi_1(U/\widetilde{H},q(u_0))=\{[\upsilon\lambda]:\lambda\text{ is a path in $U$ from $u_0$ to $h(u_0)$ for some $h\in\widetilde{H}$}\}$$This is precisely $H$, by definition of $\widetilde{H}$, and this is what we wanted.
We can also characterise: $$\deck(U/\widetilde{H};X)\cong\mathcal{N}(\widetilde{H})/\widetilde{H}$$

Is this a correct construction of the covering $p_H:(X_H,x_h)\to(X,x_0)$?
 A: It is correct using you notation $p:(U/\widetilde{H},x_h)\to(X,x_0)$ is a cover map even if $\widetilde{H}$ is not a normal subgroup of Deck(U), in general if $H<Deck(U)$ then $p_{H}:(U/{H},x_h)\to(X,x_0)$ is a cover map .In your case where $\widetilde{H}\lhd Deck(U)$ it creates a bijective correspondence between {normal cover(Galois cover) of X(except isomorphism)}<->{Normal subgroup of $\pi_1(X,x_0)$}. It is a nice way to classify normal cover of given space X when of course is  globally and locally path connected and semi-locally simply connected. Our professor explain it in an algebraic topology course. Well done have a good day
A: Another viewpoint. In a certain sense, my question is really trivial because I was taking the same quotient!
Let $X$ be a nonempty, locally and globally path connected space and let $\upsilon:(U,u_0)\to(X,x_0)$ be a based simply connected covering space of $X$.
The relation on $U$ by which Hatcher forms the space $X_H$ is precisely this: among all $\alpha,\beta\in U$, presume that they can be meaningfully identified with based homotopy classes of paths $[\gamma],[\gamma']$ that begin at $x_0$. Set $\alpha\sim\beta$ iff. $[\overline{\gamma}\cdot\gamma']\in H$. This is equivalent to saying $[\gamma]\sim[\gamma]\cdot[\eta]$ for all $[\eta]\in H$.

Note: I define path composition in what I view the most natural way, but my definition stands in contradiction to Hatcher's contravariant notation. For me, a concatenation of paths $a,b$ where $b(1)=a(0)$ shall be the path $a\cdot b$ with value at $t\in I$ defined by: $$(a\cdot b)(t)=\begin{cases}b(2t)&0\le t\le1/2\\a(2t-1)&1/2\le t\le1\end{cases}$$"First $b$, then $a$".

To check points of $U$ can be meaningfully identified with path classes in $X$, note that for any $\alpha\in U$ there is one and only one based homotopy class of paths $u_0\to\alpha$ by simple connectivity. Pick a representative $\lambda$. Then the path class $[\upsilon\lambda]$ in $X$ is well-defined as a function of $\alpha$, and it defines a path beginning at $x_0$. Conversely, if $\gamma$ is a path in $X$ beginning at $x_0$, then $\gamma$ admits a unique lift to a path $\lambda$ in $U$ that begins at $u_0$. The endpoint $\lambda(1)$ is well-defined as a function of the based homotopy class $[\gamma]$ since the choice of representative is irrelevant regarding endpoints. We assign $[\gamma]\mapsto\alpha:=\lambda(1)$. This is a mutually inverse association so is a valid way of describing elements of $U$.
Let $[\eta]\in\pi_1(X,x_0)$. There is a corresponding $\alpha\in U$ and necessarily $\upsilon(\alpha)=[\eta](1)=x_0$. By simple connectivity and inherited local path connectivity of $U$, we know that $\upsilon:(U,u_0)\to(X,x_0)$ lifts against $\upsilon:(U,\alpha)\to(X,x_0)$ to produce a map $f_{[\eta]}:U\to U$, $u_0\mapsto\alpha$. For any $\beta\in U$, corresponding to some $[\gamma]$, pick the representative $\gamma$ and let $\widetilde{\gamma}$ be the lift of $\gamma$ as a path $u_0\to\beta$: let $\widetilde{\eta}$ be the lift of $\eta$ to a path $u_0\to\alpha$.
Then $f_{[\eta]}(\widetilde{\gamma})\cdot\widetilde{\eta}$ is a path $u_0\to f_{[\eta]}(\beta)$ that projects to the path $\gamma\cdot\eta$ in $X$. Hence, under the identification of points of $U$, $f_{[\eta]}$ corresponds to taking $[\gamma]$ to $[\gamma]\cdot[\eta]$. This action is continuous and (considering $[\overline{\eta}]$ to create an inverse) is a homeomorphism over $X$, a deck transformation.
Conversely, if $g$ is a deck transformation of $U$ over $X$, let $\lambda$ represent the unique class of paths $u_0\to g(u_0)$. Then $\upsilon\lambda$ is a loop at $x_0$ in $X$, call it $\eta$. We have $g(u_0)=f_{[\eta]}(u_0)$, and because $g,f_{[\eta]}$ have the same $\upsilon$-image and agree at one point, and $U$ is connected, it follows that $g=f_{[\eta]}$ everywhere. So all deck transformations can be described in this way.
The assignment $[\eta]\mapsto f_{[\eta]}$ is a contravariant homomorphism, so we get a canonical isomorphism of groups: $$\pi_1(X,x_0)^{\mathsf{op}}\cong\operatorname{Deck}(U;X)$$
So for any $H$ a subgroup of the LHS, we get $\widetilde{H}$ a subgroup of the RHS under this isomorphism and Hatcher's $X_H$ is precisely the orbit space quotient $U/\widetilde{H}$. Our constructions are essentially the same.
