Let $p :X \to Y$ be a covering map and $Y$ path connected. Show that the cardinality of $p^{-1}(\{y\})$ is the same for every $y \in Y$. 
Let $p :X \to Y$ be a covering map and $Y$ path connected. Show that the cardinality of $p^{-1}(\{y\})$ is the same for every $y \in Y$.

Let $y_0, y_1 \in Y$, then $\exists \alpha :I \to Y$ such that $\alpha(0)=y_0$ and $\alpha(1)=y_1$. Consider now $x \in p^{-1}(\{y_0\})$. By the lifting criterion we have that $\exists \tilde{\alpha}: I \to X$ such that $\tilde{\alpha}(0)=x$. If I now define $\varphi:p^{-1}(\{y_0\}) \to p^{-1}(\{y_1\})$ and I can show that $\varphi$ is a bijection I think that would be satisfactory. What kind of choices do I have for $\varphi$ or how I should think about it?
 A: An approach that doesn't rely on path-lifting is to define an equivalence relation $y \sim y'$ iff $p^{-1}(\{y\})$ and $p^{-1}(\{y'\})$ have the same cardinality. It follows the idea given in Introduction to Topological Manifolds by John M. Lee.
Now any $y$ in $Y$ has an evenly covered neighborhood $U_y$ in $Y$. As a consequence of the definition of covering maps, we can write $$p^{-1}(U_y) = \bigcup\limits_{\alpha\in A}V_\alpha$$ where the $V_\alpha$ are disjoint (open) subsets of $X$, and $V_\alpha$ is homeomorphic to $U_y$ under the restriction $p|_{V_\alpha}: V_\alpha \to U_y$ for each $\alpha$.
Suppose $y' \in U_y$. We can define a map $f: A \to p^{-1}(\{y'\})$ by $$f(\alpha) = p|_{V_\alpha}^{-1}(y') \in p^{-1}(\{y'\}) \cap V_\alpha$$
Since the $V_\alpha$ are disjoint, $f$ must be injective. Since the $V_\alpha$ cover $p^{-1}(\{y'\})$, $f$ must be surjective. Therefore $f$ is a bijection, which shows that the cardinality of $p^{-1}(\{y'\})$ is the cardinality of $A$ for all $y' \in U_y$. This shows that $U_y$ is contained in the equivalence class of $y$, so the equivalence classes of $\sim$ are open in $Y$.
Now if $y$ is any point in $Y$, let $[y]$ be its equivalence class. As shown above it is open in $Y$. The equivalence classes partition $Y$ so the complement of $[y]$ is a union of equivalence classes, which are also open. This shows that $[y]$ is both open and closed in $Y$.
Path connectedness implies connectedness. Since $Y$ is connected and $[y]$ is nonempty, $[y] = Y$ so there is only one equivalence class.
A: Your idea is correct. Given two fibers $F_i = p^{-1}(y_i)$, you can construct a bijection $\varphi : F_0 \to F_1$ as you did:
Choose any path $\alpha : I \to Y$ such that $\alpha(i) = y_i$. For each $x \in F_0$ there exists a unique lift $\alpha_x : I \to X$ such that $\alpha_x(0) = x$. Then define $\varphi(x) = \alpha_x(1) \in F_1$. Clearly $\varphi(x)$ may depend on the choice of $\alpha$, thus one could write more precisely $\varphi_\alpha : F_0 \to F_1$.
Why is $\varphi$ a bijection? Let $\alpha^{-} : I \to Y$ be the inverse path of $\alpha$. As above this yields a function $\varphi^- : F_1 \to F_0$. Then

*

*$\varphi^-\circ  \varphi = id$:
$\varphi^-(\varphi(x))$ is obtained by taking the unique lift $\alpha^{-}_{\varphi(x)} : I \to X$ of $\alpha^{-}$ such that $\alpha^{-}_{\varphi(x)}(0) = \varphi(x)$; then $\varphi^-(\varphi(x)) =  \alpha^{-}_{\varphi(x)}(1)$. But it is clear that $\alpha^{-}_{\varphi(x)}$ is nothing else than the inverse path of $\alpha_x$, thus $\varphi^-(\varphi(x)) = x$.

*$\varphi \circ  \varphi^- = id$:
This is similar as 1.

Note that all fibers $p^{-1}(y)$ have the same cardinality under the weaker assumption that $Y$ is connected. See
If $h : Y \to X$ is a covering map and $Y$ is connected, then the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$.
