Pull back of universal cover is universal iff the map induces a fundamental group isomorphism I'm trying to solve the following task:
Let $f: Y \rightarrow X$ be a map and $p: \bar{X} \rightarrow X$ be the universal cover of $X$. The spaces Y and X are path-connected. The pull back of the cover $p$ by $f$ is a universal cover of $Y$ if and only if $f$ induces an isomorphism between the fundamental groups of $Y$ and $X$.
I think i have the easier direction. Let $Z$ be the space of the pull back and let $p' : Z \rightarrow Y$ and $p_2 : Z \rightarrow \bar{X}$ be the appropriate projections. If $f$ induces an isomorphism but $Z$ isnt simply connected, we can take a loop in $Z$ that isnt trivial. Then on one hand the composition $f \circ p'$ cannot send it to a trivial loop since it induces a monomorphism on the fundamental groups (as a composition of a cover and a function inducing an isomorphism). On the other hand the composition $p \circ p_2$ does send any loop to a trivial one, since $p$'s domain is the simply connected space $\bar{X}$. 
Is this correct? And how would i approach the other direction? I feel like this should be fairly straightforward from the definition of the pull back, but it doesn't feel that intuitive to me yet.
 A: You want to be a little careful, since the pullback $Z = Y \times_X \tilde{X}$ need not be path-connected even if $X$ and $Y$ are.  (For example, take $f: * \to S^1$.)  However, each path component of $Z$ will be simply-connected.  
I think your proof works, but here's another way to think about this problem.  Associated to a pullback square
$$
\begin{array}{ccc}
Z & \rightarrow & \tilde{X} \\
\downarrow & & \downarrow \\
Y & \xrightarrow[f]{} & X,
\end{array}
$$
there is a long exact sequence (generalizing the long exact sequence of a fibration)
$$
\cdots \to \pi_2 \tilde{X} \times \pi_2 Y \to \pi_2 X \to \pi_1 Z \to \pi_1 \tilde{X} \times \pi_1 Y \to \pi_1 X \to \cdots
$$
(This is a little sloppy since I've omitted the basepoints from the notation.)
Let's consider the first map $\pi_2 \tilde{X} \times \pi_2 Y \to \pi_2 X$.  Since the universal covering $\tilde{X} \to X$ induces an isomorphism in $\pi_n$ for $n > 1$, this map is surjective.  
Moreover, $\pi_1 \tilde{X} = 0$ by construction.  So our long exact sequence becomes 
$$
0 \to \pi_1 Z \to \pi_1 Y \xrightarrow{f_*} \pi_1 X \to \cdots
$$
So assuming I haven't made a mistake, we deduce that $\pi_1 Z = 0$ iff $f_*: \pi_1 Y \to \pi_1 X$ is injective.  
