Covering map of quotient is Galois iff subgroup is normal I am having difficulty with part of the following exercise from Forster's Lectures on Riemann Surfaces:

Let $X$ be a connected manifold and $p:\tilde{X}\rightarrow X$ be its universal covering. Let $G\leq Deck(\tilde{X}/X)$ be a subgroup and $Y$ be the quotient of $\tilde{X}$ by the equivalence relation $x\sim y$ if $g(x)=y$ for some $g\in G$, and let $q: Y\rightarrow X$ be the map induced by $p$. Show that $q$ is a covering map which is Galois if and only if $G$ is a normal subgroup of $Deck(\tilde{X}/X)$.

As far as I can tell, showing that $q$ is a covering map is fairly straightforward and does not depend on $G$ being normal (right?). I believe I can show that if $G$ is normal then the covering is Galois as follows:
Letting $[x],[y]\in\tilde{X}$ such that $q([x])=q([y])$ (then also $p(x)=p(y)$) there exists $\sigma\in Deck(\tilde{X}/X)$ such that $\sigma(x)=y$ because universal coverings are Galois. Given $x'\in [x]$ so that $g(x)=x'$ for some $g\in G$, 
$$\sigma(x')=(\sigma\circ g) (x)=(g'\circ\sigma)(x)=g'(y)$$
so that $\sigma(x)\sim\sigma(x')$. Thus, $\sigma$ induces a mapping on $Y$ that sends $[x]$ to $[y]$ so $q$ is Galois.  
I am unsure, however, how to prove the reverse implication. My thought is to look at the homomorphism $Deck(\tilde{X}/X)\rightarrow Deck(Y/X)$ by looking at the action of $\sigma\in Deck(\tilde{X}/X)$ on elements of $Y$ as above and then argue that the kernel is precisely $G$ but the problem is without the assumption of normality it's unclear that this map is well-defined.
 A: I have been confused by this question for 4 days and nights (and exactly from the same text book), now I'd glad to share my answer.
Your thoughts of considering a homomorphism $$\phi:\mathrm{Deck}\left(X'/G\big/X\right)\longrightarrow\mathrm{Deck}(X'/X)$$ is right.
For any $\sigma\in \mathrm{Deck}(X'/X)$, and any $x\in X'$, it is clearly that $[x],[\sigma(x)]$ are lying in a same fiber of $q$, thus by the Galois covering, there exists a unique (the uniqueness comes from the connectedness of $X$, see Remark of 5.5 in page 34 of Foster's book) Deck transformation $f_\sigma:X'/G\longrightarrow X'/G$ such that $f_\sigma([x])=[\sigma (x)]$, thus the map
$$\phi: \sigma\mapsto f_\sigma$$
is a well-defineded homomorphism, moreover we have $\mathrm{ker}\phi=G$, thus a normal subgroup.
Moreover, we can know from the connectedness that the universal covering $X'$ is actually simple connected, thus $X'/G$ is at least a connected space, besides, the usual quatient map $\pi: X'\longrightarrow X'/G$ is also a covering map (hence also Galois), thus by theorem 5.9 of Foster's book, we have $$\pi_1(X'/G)\cong G$$
Which is also a normal subgroup of $\pi_1(X)\cong \mathrm{Deck}(X'/X)$
