Prove if $p:\tilde{M} \to M$ is covering and $\tilde{M}$ is irreducible then $M$ is also I was tring to understand the proof for the fact that for 3-manifold, if $p:\tilde{M} \to M$ is covering and $\tilde{M}$ is irreducible then $M$ is also irreducible, there is a proof in Hatcher's 3-manifold notes in page 12 as follows:
Proof: A sphere $S \subset M$ lifts to spheres $\tilde{S} \subset \widetilde{M}$. Each of these lifts bounds a ball in $\widetilde{M}$ since $\widetilde{M}$ is irreducible. Choose a lift $\tilde{S}$ bounding a ball $B$ in $\widetilde{M}$ such that no other lifts of $S$ lie in $B$, i.e., $\tilde{S}$ is an innermost lift. We claim that $p: B \rightarrow p(B)$ is a covering space .....
The question is what $\tilde{S}$ be the innermost lift means?why there is some innermost lift, I can't get the idea.
 A: Let $f : S^2 \rightarrow M$ be the embedded sphere, suppose there is no inner most lift, then there is a sequence of lifts of $f$, $g_k$ $k=1...$, such that  each $g_k(S^2) $ is contained in the ball bounded by $g_{k-1}$ and all these lifts are distinct. Then there is some point $x \in \tilde{M}$ contained in the closures of all the balls, such that $x$ is a limit point of the images of the sequence of lifts, i.e. WLOG we can assume $x = \lim_{k} g_k(t_k)$ for some $t_k \in S^2$.
Now let $U$ be a path connected neighbourhood of $x$, such that $p: U \rightarrow p(U)$ is a homeomorphism (diffeomorphism) onto its image. There exist points $s_k \in g_k(S^2) \cap U $, $s_k = g_k(t_k)$ converging to $x$, by continuity of $p$ we have $p(s_k) = f(t_k)$ converging to $p(x)$, passing to a subsequence we may assume $t_k$ also converges to some $y \in S^2$, so that $f(y) = p(x) = \lim_{k} p(s_k)$.
But this means there is some lift of $f$ sending $y$ to $x$, call it $g_0$. Take a path $\gamma_k$ in a path connected neighbourhood $V$ of $f(S^2) \cap p(U)$ of $f(y) \in f(S^2) \cap p(U)$ from $f(y)$ to some point $f(t_k) \in V$ where $k$ is chosen sufficiently large so that $f(t_k) \in p(U)$, this is necessarily given by $f(\gamma'_k)$ for some path $\gamma'_k$ in $S^2$ from $y$ to $t_k$ (since $f$ is a homeomorphism),now  $\gamma_k$ lifts uniquely to a path in $U \subset \tilde{M}$ starting at $x$, and we know that this path is given by $g_0(\gamma'_k)$ since $p \circ g_0 = f$ and $g_0(y) = x$. But this means that $g_0(t_k) = g_k(t_k)$ since both are mapped by $p$ to $f(t_k)$ and both are contained in $U$, where $p$ is a homeomorphism (in particular, where $p$ is injective). Hence for all $k$ sufficiently large, $g_0$ and $g_k$ coincide at some point, but since these are both lifts of $f$, they must be equal for $k$ sufficiently large, which is a contradiction since we assume all the $g_k$ are different lifts of $f$.
Edit: I think my argument I stated earlier was not correct and basically an oversimplification, the whole images $g_k(S^2)$ need not be contained in some neighbourhood of $x$, I have tried to amend it and hopefully it now goes through!
