Covering map between two path connected sets First off, I see a lot of variations of this problem cropping up on practice qualifiers, and I'm trying to regain my knowledge of topology.
Let $p: X \to Y$ be a covering map where $X$ and $Y$ are path connected.  Suppose that for any two points $y_1,y_2\in Y$ such that $p^{-1}(y_1)$ and $p^{-1}(y_2)$ are finite.  Then $p^{-1}(y_1)$ and $p^{-1}(y_2)$ have the same cardinality.
Since $X$ is path connected we know that $\pi(X,p^{-1}(y_1))$ and $\pi(X,p^{-1}(y_2))$ are isomorphic (and hence one-to-one).  Is this the whole argument?  I'm not using any of the assumptions placed upon $Y$.
 A: This result is true under weaker hypothesis - connectedness (not necessarily path connectedness) of $Y$ is all we need.
Fix a point $y_1\in Y$. Let $|p^{-1}(y_1)|=k$. Suppose  there is some $y_2\in Y$ such that $|p^{-1}(y_2)|\neq k$
Define the set $A:=\{y\in Y\ |\ |p^{-1}(y)|=k\}$. We would like $A$ to be all of $Y$. For this we will use the connectedness of $Y$.
For this we will show that both $A$ and $A^c$ are open.
To show $A$ is open -
Let $y\in A$. Let $U$ be an evenly covered neighbourhood of $y$ in $Y$. We claim that $U\subseteq A$ which will show that $A$ is open. 
Proof of claim : Let $z\in U$. Now since $y\in U$ and it is evenly covered, we can write $p^{-1}(U)=\bigsqcup_{i=1}^{k} V_i$ such that Each $V_i$ is open in $X$ and  $p|_{V_i}:V_i\to U$ is a homeomorphism. So there is exactly one preimage of $z$ in each $V_i$. Thus $|p^{-1}(z)|=k$. 
To show that $A^c$ is open -
Let $z\in A^c$ and let $|p^{-1}(z)|=l\neq k$. Let $W$ be an evenly covered neighbourhood of $z$. the a similar argument as above will show that $W\subseteq A^c$, proving it is open.
Conclusion -
Clearly $y_1\in A$ and $y_2\in A^c$ so that they are non empty and (as seen above) open. This forms a partition of $Y$ contradicting its connectedness. Thus $Y=A$ and you are done.
A: Here is a proof:
Let $\gamma$ be a path from $y_1$ to $y_2$.  For each point $x \in p^{-1}(y_1)$, there is a unique lift $\gamma_x$ of $\gamma$ to a path which begins and $x$ and ends at a point $x'$ of $p^{-1}(y_2)$.  The map $x \mapsto x'$ from $p^{-1}(y_1)$ to $p^{-1}(y_2)$ is an injection: two of these lifts can't intersect, and therefore can't end at the same point.  If they did intersect then the covering map would not be a local injection in a small neighborhood of an intersection point.  Now reverse the roles of $y_1$ and $y_2$ to get an injection in the other direction.  Since the two sets are finite, this shows that they have the same size.
