simply connected covering of a path connected space (II) Let $p:\overline{X}\rightarrow X$ be a simply connected covering of a path connected space $X$ and $A\subset X$ be a path connected set.

Show that the inclusion induced homomorphism $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is injective iff each path component of  $p^{-1}(A)$ is simply connected.

First of all i do not think i understand what does it mean to say $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is injective..

Is it like if i have a loop in $A$ which is not nullhomotopic in $A$ (existence of $H:I\times I\rightarrow A$) then it is not nullhomotopic (existence of $H:I\times I\rightarrow X$) in $X$.

Let me know if this is what it actually mean...

Assume $i_{\sharp}$ is injective and let $\omega$ be a loop in $p^{-1}(A)$... See this as a loop in $\overline{X}$... as $\overline{X}$ is simply connected $\omega$ is nullhomotopic in $\overline{X}$ i.e., we have 
$H:I\times I\rightarrow \overline{X}$ such that 

$H(t,o)=\omega(t)$ and $H(t,1)=w(0)$ for all $t\in I$..

Compose this with $p$ to get $I\times I\xrightarrow{p\circ H}X$ with $(p\circ H)(t,0)=(p\circ \omega)(t)$ and $(p\circ H)(t,1)=(p\circ \omega)(0)$.. 

So, this $p\circ \omega$ is null homotopic in $X$ so it has to be null homotopic in $A$ as well.. As $p\circ \omega$ is nullhomotopic in $A$ i belive this would imply $\omega$ is nullhomotopic in $p^{-1}(A)$.. 

I could not think of any ideas about converse part and how to prove if $p^{-1}(A)$ is not actually path connected...

Please give only hints..
You are right about the first part. If $p\omega$ is nullhomotopic in $A$, then the homotopy to the constant loop lifts to a homotopy between $\omega$ and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map $p:\tilde X\to X$ restricts to a covering map $p^{-1}(A)\to A$ for every subset $A$ of $X$.

Regarding the second part, let every path component of $p^{-1}(A)$ be simply-connected.  If $\omega$ is a loop in $A$, nullhomotopic as a loop in $X$, then what can you say about its lift $\tilde\omega$. Recall that every homotopy of paths lifts to a homotopy between their unique lifts at a given point $\tilde x$ lifting the starting point of the paths.
The Statement is not true. Für example $X=\overline{X}={\mathbb R}^2$, which is simply connected, p the Identity, and $A=S^1$. The Fundamental Group of A is nontrivial, so $i_\sharp$ is not injective, but $p^{-1}A=A$ is connected.

By the way, the Statement as above has no meaning anyway. The Path component is path connected anyway, you must mean something else.