Each slice is mapped homeomorphically onto a space by the covering map $p$ I am reading this proof, but I don't understand what's intention of saying each $E_0\cap V_\alpha$ is mapped homeomorphically onto $U\cap B_0$ by $p$ (the last sentence). Does it imply that $p|E_0$ is surjective by the bijection property? 
The definition I use for covering map is that if $p:E\to B$ is continuous and surjective, and for any $b\in B$, there exists a nbhd $U$ of $b$ in $B$ which is evenly covered by $p$, then $p$ is a covering map.



 A: We have that for each $\alpha$, $p: V_\alpha \to U$ is a homeomorphism.
Note that $U \cap B_0$ is open in $B_0$ and its inverse image is $$p^{-1}[U \cap B_0] = p^{-1}[B_0] \cap p^{-1}[U] = E_0 \cap (\bigcup_\alpha V_\alpha) = \bigcup_{\alpha} (V_\alpha \cap E_0)$$
and the latter is the partition of this inverse image into disjoint open sets.
We also have that $$p[V_\alpha \cap E_0] = U \cap B_0$$ 
(if $x \in V_\alpha \cap E_0$ then $p(x) \in p[U] = V$ and $p(x) \in B_0$, as $x \in E_0 = p^{-1}[B_0]$, which shows the left ro right inclusion; if $y \in U \cap B_0$ we find $x \in V_\alpha$ with $p(x) = y$ as $p: V_\alpha \to U$ is surjective and by definition this $x \in p^{-1}[B_0] = E_0$ as well, so $y \in p[V_\alpha \cap E_0]$ showing the other inclusion.)
We just restrict $p$ on both sides and we still have a homeomorphism between $V_\alpha \cap E_0$ and $U \cap B_0$ for each $\alpha$ (restrictions of continuous maps are continuous; also applies to the inverse of $p$). So $U \cap B_0$ is the required evenly covered neighbourhood of $b_0$.
