Suppose that we fix a connected locally path connected space $X$ and two covering projections $p_i: X_i\to X$, $i=1,2$ with $X_i$ connected, and $f:X_1\to X_2$ is a continuous map making the obvious diagram commute. Then $f$ is a covering projection if $f$ is surjective, so all you have to prove is that this holds assuming $X$ is connected locally connected. Recall that if $X$ is locally path connected so are the $X_i$, and since they are connected by hypothesis, they are path connected. Given $x_2\in X_2$, take $x_1\in X_1$ and a path $\omega_2$ from $f(x_1)$ to $x_2$. Consider now the path $p_2\omega_2$ in $X$. Since $p_1$ is a covering, there is a path $\omega_1$ starting at $x_1$ in $X_1$ such that $p_1\omega_1=p_2\omega_2$. But then $p_1=p_2f$ gives $f\omega_1=\omega_2$ by the unique lifting property. In particular $f\omega_1(1)=\omega_2(1)=x_2$, so $f$ is onto.
To each path-connected covering space $p:(\tilde{X},\tilde{x}_0)\to (X,x_0)$ assign the subgroup $G(p) = p_*(\pi_1(\tilde{X},\tilde{x}_0))$ of $\pi_1(X,x_0)$. Then the assertion is
Each subgroup $G \subset \pi_1(X,x_0)$ has the form $G = G(p)$ for some path-connected covering space $p$.
$p,p'$ are basepoint-preserving isomorphic covering spaces if and only if $G(p) = G(p')$.
$p,p'$ are isomorphic covering spaces if and only if $G(p),G(p')$ are conjugate.
In 3. we only require that there exists a homeomorphism $f : \tilde{X} \to \tilde{X}'$ such that $f \circ p = p'$; we do not require $f(\tilde{x}_0) = \tilde{x}'_0$.
You consider a fixed covering space $p_1:(X_1,x_1)\to (X,x_0)$ and define $Q(X_1,x_1)$ as the collection of all covering spaces $p_2:(X_2,x_2)\to (X,x_0)$ where there is a basepoint-preserving homeomorphism $f:X_1\to X_2$ with $p_1=p_2\circ f$. If you drop the reqirement "basepoint-preserving", you get a larger class $Q'(X_1,x_1)$. If you replace $x_1$ by $x'_1$ in the same fiber, you trivially have $(X_1,x'_1) \in Q'(X_1,x_1)$, but in general not $(X_1,x'_1) \in Q(X_1,x_1)$. If $p_1$ is a normal covering space, then it is true for all $x'_1$. Thus it is possible. But if $G(p_1)$ is a non-normal subgroup of $\pi_1(X,x_0)$, you will find $x'_1$ such that $(X_1,x'_1) \notin Q(X_1,x_1)$.
Best Answer
Yes. Doing it this way allows for a cleaner statement of the Galois correspondence, which naturally generalizes even to the case that $X$ is not assumed to be path-connected:
The connected covering spaces correspond to the transitive $\pi_1(X)$-sets, which correspond to conjugacy classes of subgroups. But this way of stating the Galois correspondence tells you not only what non-connected covering spaces look like but also what maps between covering spaces look like, in a nice clean way. The resulting category is also much more nicely behaved than if we restrict to the connected case, e.g. it now has products and coproducts.
To generalize to the case that $X$ is not path-connected we replace the fundamental group $\pi_1(X)$ with the fundamental groupoid $\Pi_1(X)$.