Prove $p:E\to B$ is a homeomorphism Let $p:E\to B$ be a covering projection.
$E$ is path connected, $B$ is simply connected. 
Prove $p:E\to B$ is a homeomorphism (bijective surjective, continuous and open).

I don't understand something basic... 
It seems to me as if it's always true regardless of whether $B$ is simply connected.
Observe the preimage $V=p^{-1}(B)\subseteq E$. 
from the definition of $p$ follows: $B$ is open hence $V$ is open and is the union of disjoint open sets $V=\cup V_\alpha$.
Now $p^{-1}(B)=\{e\in E | p(e)\in B\}=E$, hence $V=E$.
But $E$ is path connected hence connected, so it cannot be the union of disjoint open sets, so there must be a single set in the union $\{V_\alpha\}_\alpha=\{V\}$. and $p|_{V_\alpha}=p|_V = p|_E = p:E\to B$ is homeomorphism.
Where did I go wrong?
Does the definition of $p$ holds only for proper subsets of B?
If you can, please put me in the right direction :)
 A: The definition of $p: E \to B$ being a covering map:

Each point $x \in E$ has an open neighborhood $V$ whose preimage is a disjoint union of open sets $W_\alpha$ such that each $p: W _\alpha \to V$ is a homeomorphism.

You seem to have chosen $V = B$. You then show this implies $W_\alpha=E$ and so $p$ is a homeomorphism. 
But according to the definition you do not get to choose what $W_\alpha$ looks like. You only know it exists. That's the problem.
A: The proof you gave is wrong as Daron showed.
But theorem 54.4 in Munkres (2nd edition) says :

let $p:E \to B$ be a covering map. Let $p(e_0) = b_0$. If $E$ is path connected, then the lifting correspondence (defined the paragraph above this theorem )
  $$\pi_1(B, b_0) \to p^{-1}[\{b_0\}]$$ is a surjection. 

Now, $B$ is simply connected so for any $b_0 \in B$, $|\pi_1(B, b_0)|  =1$, and this theorem implies then that all $p^{-1}[\{b_0\}]$ are singletons too. 
So $p$ is 1-1 and all covering maps are surjective, open (even local homeomorphisms) and continuous. So $p$ is a homeomorphism.
