Show that the index is 2 if and only if $\tilde{X}$ is path-connected. Show that the index is 1 if and only if $\tilde{X} \cong X\sqcup X$ 
Let $\pi\colon \tilde{X} \to X$ be a double cover, and let $X$ be path-connected. Show that the index is 2 if and only if $\tilde{X}$ is path-connected. Show that the index is 1 if and only if $\tilde{X} \cong X\sqcup X$.

Let $G = \pi_1(X, x_0)$ and $H = \pi_*(\pi_1(\tilde{X}, \tilde{x}_0))$. If indeed $\tilde{X} = X \sqcup X$, it is clear that $\pi_1(\tilde{X}, \tilde{x}_0)$ is precisely $G$. Also, if $\tilde{X}$ happens to be path connected, then it is possible to choose a path $\tilde{\gamma}$ in $\tilde{X}$, joining the two pre-images of $x_0$ under $\pi$, say $\tilde{x}_0$ and $\tilde{x}_1$. Then, $\gamma = \pi\circ \tilde{\gamma}$ is a loop in $X$ based at $x_0$; since its lift $\tilde{\gamma}$ is not a based loop in $\tilde{X}$, we have $H \neq G$.
However, I am not sure how to show the reverse implication in the second part, nor am I able to figure out the details in the first.
 A: It is well-known that the number of sheets of a covering map $p : \tilde X \to X$ with $X$ and $\tilde X$ path-connected equals the index of $p_*(\pi_1(\tilde X,\tilde x_0))$ in $\pi_1(X,x_0)$, where $\tilde x_0 \in \tilde X$ is arbitrary and $x_0 = p(\tilde x_0)$.

*

*If $\tilde X$ is path connected, then the index is $2$.

This follows from the above theorem.


*If $\tilde X$ is not path connected, then the index is $1$.

Proof.
(i) Let $C$ be a path component of $\tilde X$. Then $C \cap p^{-1}(x) \ne \emptyset$ for all $x \in X$.
Pick $\tilde x^* \in C$ and let $x^* = p(\tilde x^*)$. Then $C \cap p^{-1}(x^*) \ne \emptyset$. Let $x \in X$ and $u$ be a path in $X$ such that $u(0) = x^*$ and $u(1) = x$. Lift $u$ to a path $\tilde u$ in $\tilde X$ such that $\tilde u(0) = \tilde x^*$. Then $\tilde x = \tilde u(1) \in C \cap p^{-1}(x)$.
(ii) $\tilde X$ has exactly two path components $C_1, C_2$; they have the property that $C_i \cap p^{-1}(x)$ is a singleton for all $x \in X$.
Let $\{C_\alpha\}_{\alpha \in A}$ be the set of all path components of $\tilde X$. Then all $C_\alpha \cap p^{-1}(x) \ne \emptyset$ and $p^{-1}(x)$ is the disjoint union of these non-empty sets. Since $p^{-1}(x)$ has two elements, the above claim follows.
(iii)  $p_*(\pi_1(\tilde X,\tilde x_0)) = \pi_1(X,x_0)$ which means that the index is $1$.
Let $[u] \in \pi_1(X,x_0)$. The closed path $u$ in $X$ lifts to a path $\tilde U$ in $\tilde X$ such that $\tilde u(0) = \tilde x_0$. Since the image of $\tilde u$ is contained in the path component $C$ of $\tilde x_0$, we have $\tilde u(1) \in C \cap p^{-1}(x_0) = \tilde x_0$. Thus $\tilde u$ is a closed path and by construction $p_*([\tilde u]) = [u]$.
From 1. and 2. we infer that


*The index is $2$ if and only if $\tilde X$ is path-connected.


*The index is $1$ if and only if $\tilde X$ is not path-connected.
Therefore the statement


*The index is $1$ if and only if $\tilde X \cong X\sqcup X$
is equivalent to


*$\tilde X$ is not path-connected if and only if $\tilde X \cong X\sqcup X$.

The if-part is trivially true, but there are examples where $\tilde X$ is not path-connected, but does not have the form $\tilde X \cong X\sqcup X$ (see below). Thus 6. is false and a fortiori 5. is false. Instead of 6. we have


*The following are equivalent:
(a) $\tilde X \cong X\sqcup X$.
(b) $\tilde X$ has two path components $C_1, C_2$ which are open in $\tilde X$.
(c) $\tilde X$ is not connected.

Proof.
(a) $\implies$ (c): Trivial.
(c) $\implies$ (b): We have $\tilde X = U_1 \cup U_2$ with disjoint nonempty open $U_1, U_2 \subset \tilde X$. Moreover, since $\tilde X$ cannot be path-connected, it has two path components $C_1, C_2$. The $C_i$ are connected, thus cannot intersect both of the $U_j$. Hence both $C_i$ must be contained in some $U_{j(i)}$. Clearly $j(1) \ne j(2)$ since $C_1 \cup C_2 = \tilde X$. Therefore $C_i = U_{j(i)}$.
(b) $\implies$ (a): $X$ is the union of the two disjoint open subspaces $C_i$. It therefore suffices to show that the $p_i  = p\mid_{C_i} : C_i \to X$ are homeomorphisms. It follows from (ii) that the $p_i$ are continuous bijections. Since $p$ is an open maps, also their restrictions $p_i$ to the open subspaces $C_i$ are open maps. Thus the $p_i$ are homeomorphisms.
A well-known condition assuring that the path components of $\tilde X$ are open in $\tilde X$ is that $\tilde X$ is locally path-connected (which is equivalent to $X$ being locally path-connected because $p$ is a local homeomorphism).
Thus 5. is true for locally path-connected $X$.
Remark. In 7.(b) it suffices to require that one of the $C_i$ is open. In fact, it was shown in Does there exist a double cover with trivial deck transformation group? that there exists a deck transformation $h: \tilde X \to \tilde X$ flipping the points in all fibers $p^{-1}(x)$. We therefore get $h(C_1) = C_2$ which shows that if one of the $C_i$ is open, then also the other is open. Moreover it shows that $C_1, C_2$ are always homeomorphic.
Here is the promised example of a non-path-connected $\tilde X$ which does not split as $\tilde X \cong X\sqcup X$.
Let $X$ be the Warsaw circle. This is a path-connected compact subspace of the plane. Many questions in this forum deal with it. A double covering of $X$ can easily be obtained as follows. Regard $X$ as a subset of $\mathbb C^* = \mathbb C \setminus \{0\}$ such that $0$ is contained in the bounded open region enclosed by $X$. The map $f : \mathbb C^* \to \mathbb C^*, f(z) = z^2$, is a covering map with two sheets. The restriction
$$p = f \mid_{f^{-1}(X)} : \tilde X = f^{-1}(X) \to X$$
is again a covering map with two sheets. The space $\tilde X$ is connected, thus it does not satisfy $\tilde X \cong X\sqcup X$.
