Property of a space induced by fundamental group and covering space. This problem is from my past Qual.
Let $(X,Y)$ be a CW pair with both $X,Y$ connected and $x_0\in
  X$ a basepoint. Assume that inclusion induced homomorphism
  $\pi_1(Y,x_0)\to\pi_1(X,x_0)=G$ is injective and denote its image by $H\leq G$.
  Let $p:(\tilde X,\tilde x_0)\to (X,x_0)$ be the
  universal cover and let $p_H:(\tilde X_H,\tilde x_H)\to (X,x_0)$ be
  the covering corresponding to $H$.
1)  Prove that each component of $p^{-1}(Y)$ is
  simply connected (and hence is a universal cover of $Y$).
2) Show that there is a natural bijection between the cosets $gH$
  of $H$ in $G$ and the components of $p^{-1}(Y)$.
I learned algebraic topology by Hatcher's book. Anyway, the theorems in Section 1.3 (Covering space) in the book are mostly about giving a space and give some properties about the fundamental groups and its bijection with the covering space.
This is quite the opposite. It gives me some properties about the fundamental group and tell us to go back to the space. So I don't know how to approach. Like how to prove a space is simply connected? The definition will lead me to point-set topology, which I am not good at.
I feel the problem harder since it gives me a CW pair. While section 1.3 in the book deals with general spaces. So I think I need something from CW complex here.
THank you.
 A: This is a nice question. The following is an argument for the first part. I'll assume that $\tilde{X}$ is connected since you called it the universal cover.
Let $\tilde{Y}=p^{-1}(Y)$. Then, it is clear that $q=p \mid_{\tilde{Y}}$ is also a covering map. Moreover, $q_*$ is injective. Now, let $i_*:\pi_1(Y) \to \pi_1(X)$ be induced by inclusion, and let $\tilde{i}$ be its lift. Then there is an equality
$i*\circ q_*= p_* \circ \tilde{i}_* $,
but since $i_* \circ q_*$ is injective, we can deduce that $\tilde{i}_*$ is as well, so $\pi_1(\tilde{Y})=0$.
I've omitted basepoints everywhere but this works on each component of the preimage. 

The previous argument can be strengthened if the inclusion map gives an isomorphism on the fundamental group.
If $(K,L)$ is a $CW$ pair such that $i_*:\pi_1(L) \to \pi_1(K)$ is an isomorphism, and $\tilde{K}$ is the universal cover of $K$, then we can strengthen the last argument to show that $\tilde{L}$ is connected. Indeed, we know that $p_i(\tilde{K},\tilde{L}) \cong \pi_i(K,L)$ for $i \geq 1$ by the homotopy lifting property, and by the LES of a pair, we have that
$$\pi_1(L) \to \pi_1(K) \to \pi_1(K,L) \to \pi_0(L) \to \pi_0(K)$$
is exact, but by our assumption on connectedness, the first and last maps are isomorphisms, from which we can deduce that $\pi_1(\tilde{K},\tilde{L})=0$, so applying this sequence again to the covers, we see that
$$0 \to \pi_0(\tilde{L}) \to \pi_0(\tilde{K}) $$
is exact (When interpreted correctly), so $\tilde{L}$ is connected.

The reason I went on my tangent in the middle was because I have a proof idea for (2) that I will try to flesh out, but I have a strong feeling that I'm being an idiot here.
We know that $i_*:\pi_1(Y,y_0) \subset \pi_1(X,x_0)$, so there is a corresponding (Connected) cover $r:X_1 \to X$ such that $\pi_1(X_1,x_1)=i_*(\pi_1(Y,y))$. 
It can be shown that right cosets of $r_*(\pi_1(X_1,x_1))=i_*\pi_1(Y,y_0)$ are in bijection with lifts of $x_0$ in $X_1$ (see prop VI.19 here .)
Moreover, we know that there is a lift $\tilde{r}:(Y,y_0) \to (X_1,x_1)$ of $i:(Y,y_0) \to (X,x_0)$ that is an isomorphism on $\pi_1$ with the image connected, and hence $((X_1,x_1),r(Y,y_0))$ satisfies the conditions of the middle paragraph so the lift of $r(Y,y_0))$ is connected in $\tilde{X}$. 
I hope that this can yield a proof with maybe some more effort. The idea is that we have reduced the problem to showing that preimages of basepoints in $X_1$ correspond to connected components of $\tilde{Y} \subset \tilde{X}$.
