In Hatcher's Algebraic Topology Proposition 1.40, it is said that, for a covering action of a group $G$ on a space $Y$,
(b) $G$ is the group of deck transformations of the covering map $Y\to Y/G$ if $Y$ is path-connected.
It seems to me that $Y$ connected is sufficient, since the unique lifting property only required connectedness. Am I right?
(c) $G$ is isomorphic to $\pi_1(Y/G)/p_*(\pi_1(Y))$ if $Y$ is path-connected and locally path-connected, but it seems to me by looking closely at the proof that $Y$ path-connected is sufficient.
Indeed the full power of the lifting criteria does not seem needed to build the deck transformation from the action of $g\in G$, since only loops are concerned. Am I right?
Edit Following the comments of @FShrike, here is my approach when I only suppose $Y$ path-connected.
(i) Let the covering map be $p: Y\to Y/G$. Choose $y_0\in Y$ and $b_0=p(y_0)\in Y/G$. Let $\gamma:I\to X/G$ be a loop with $\gamma(0)=\gamma(1)=b_0$
(ii) By the path-lifting theorem there exists a unique path $\widetilde{\gamma}: I\to X$ with $\widetilde{\gamma}(0)=y_0$ and $p\circ\widetilde{\gamma}=\gamma$. Since $p(\widetilde{\gamma}(0))=p(\widetilde{\gamma}(1))$, then there is and only one because of unique lifting property $g\in G$ with
$$\widetilde{\gamma}(1)=g.y_0$$
It is easy to check that such $g$ does not depend of the choice of $\gamma$ inside the homotopy class of $\gamma$ in $\pi_1(Y/G, b_0)$. Therefore we get a well defined map
$$\pi: \pi_1(Y/G, b_0)\to G$$
(iii) It is easy to check that this map is a morphism of groups and that its kernel is the isotropy group of the monodromy action $p_*(\pi_1(Y, y_0))$.
(iv) Now to show that this map is surjective is where the path-connected hypothesis on $Y$ is used: let $g\in G$. There exists a path $\sigma: I\to Y$ with $\sigma(0)=y_0$ and $\sigma(1)=g.y_0$, hence
$$g=\pi(p\circ\sigma)$$
Quite straightforward proof if I am not wrong.
Edit 2 In his proof of (c) of Proposition 1.40, Prof. Hatcher uses the proof of (b) of proposition 1.39, which uses the lifting criterion 1.33 whose proof indeed requires local path-connectedness hypothesis.
This is a bit of an overkill in my opinion, since proposition 1.39 is about proving the same thing as proposition 1.40 but in the case where $G$ is supposed to be the group of deck transformations $G(Y)$.
To show that $G(Y)$ is actually a covering action, we really need the hypothesis of local path-connectedness to be able to use the lifting criterion.
But in proposal 1.40 we already begin with a group $G$ having a covering action on $Y$!, and so its results just come from algebraic view, no need for local path-connectedness.
Best Answer
$\newcommand{\N}{\mathcal{N}}\newcommand{\G}{\mathcal{G}}\newcommand{\deck}{\operatorname{Deck}}\newcommand{\X}{\widetilde{X}}\newcommand{\x}{\widetilde{x}}\newcommand{\op}{{^\mathsf{op}}}$Let's assemble the necessary ingredients. What I will do is go through my notes and point out exactly where each hypothesis is used for each "ingredient". I'd like to note that I write the group action differently to Hatcher; for me, if $f,g$ are homotopy path classes, $fg$ denotes the homotopy path class of the composite path $a\overset{g}{\to}b\overset{f}{\to}c$.
A lot of this comes down to how one defines "normal". I've written this partially in response to a request so that this question gets properly answered and I think it's good to have the hypotheses clearly pointed out.
By proposition $1.34$, for a connected covering $p:\X\to X$ the elements of $\deck(\X;X)$ are completely determined by their values at a single point.
Hatcher defines the covering to be a normal covering if for all $x\in X$, $\x_1,\x_2\in p^{-1}(x)$ there exists $\tau\in\deck(\X;X)$ with $\tau(\x_1)=\x_2$. If the covering is normal and connected, there is a unique such $\tau$.
If the base space $X$ is globally and locally path connected, connected coverings $p:\X\to X$ are normal if and only if $H:=p_\ast\pi_1(\X;\x_0)$ is a normal subgroup of $G:=\pi_1(X;p(\x_0))$ for some $\x_0\in\X$ if and only if this is true for all $\x_0\in\X$. Write $x_0:=p(\x_0)$.
In conclusion, under no other hypotheses at all, a normal cover $p:\X\to X$ implies that the subgroups $H$ are always normal in $G$, but the local path connectivity of $X$ and global path connectivity of $\X$ are crucial in obtaining the converse implication.
We also would like to prove $\N(H)/H\cong\deck(\X;X)\op$ holds - but when? We may drop the contravariance in "$\mathsf{op}$" because every group is isomorphic to its dual, but my preferred conventions make the contravariance the natural choice.
We put $\Phi:\N(H)\to\deck(\X;X)\op$ by assigning to every $g$ its unique lifted homotopy path class $[\gamma]:\x_0\to\x_1$, and $g\in\N(H)$ means $\x_1$ must be such that $p_\ast(\pi_1(\X;\x_0))=p_\ast(\pi_1(\X;\x_1))$, so under the full hypotheses of local path connectivity of $X$ and connectivity of $\X$ there is a unique deck transformation taking $\x_0$ to $\x_1$; we define $\Phi(g)$ to be this unique deck transformation. It's then not too hard, assuming path connectivity of $\X$, to check $\Phi$ is a surjective homomorphism. The kernel of $\Phi$ is always $H$ (if $\Phi$ is well-defined, that is) so the first isomorphism theorem kicks in.
If $\N(H)=G$ i.e. $H$ is normal in $G$ but we don't already know the cover is normal, we again need the local path connectivity hypothesis to define $\Phi$. But if the cover is given to be normal, then $\N(H)=G$ is true without any other hypothesis and $\Phi$ is well-defined without any need for local path connectivity of $X$, so $G/H$ embeds as a subgroup of $\deck(\X;X)\op$; if $\X$ is also assumed to be path connected, we get the full isomorphism $G/H\cong\deck(\X;X)\op$.
In conclusion, $G/H\cong\deck(\X;X)\op$ is always true if the cover is path connected and normal in the sense of Hatcher. If the cover is only normal in the sense that $H$ is normal in $G$, then we need $X$ to be locally path connected to make this work.
In the case of (left) group covering actions of $\G$ on a space $Y$, it's always true that $q:Y\twoheadrightarrow Y/\G$ is a normal covering. If $Y$ is connected, you were quite right in saying that $\G\cong\deck(Y;Y/\G)$ follows (if $Y$ is not connected we can at least say $\G$ embeds as a subgroup of $\deck(Y;Y/\G)$).
Now for the main question - when can we say $\G\op\cong\pi_1(Y/\G;q(y_0))/q_\ast(\pi_1(Y;y_0))$? ($=G/H$) We know the cover is normal. Therefore, we need only have $Y$ path connected to make everything work. If we don't even assume $Y$ is path connected, we can just conclude that $\pi_1(Y/\G;q(y_0))/q_\ast(\pi_1(Y;y_0))$ embeds as a subgroup of $\G\op$.
The local path connectivity of $X$ is only needed in the instance that $\N(H)\neq G$ or that $\N(H)=G$ but we haven't, a priori, been told that the cover is a normal cover.