Euler characteristic of covering space of CW complex I am trying to prove the following statement: if $X$ is a finite CW complex and if $Y \to X$ is a $n$-sheeted covering then $Y$ is a finite CW complex and $\chi(Y)=n \cdot \chi(X)$. 
I know the covering space of CW complex is again a CW complex, but I am just not sure how to prove if I have one $k$-cell, the covering space have $n$ $k$-cells. The $0$-cell case is easy. For the $1$-cell of covering space, we can determine it by considering how $1$-cell of the CW complex attach to the points. But what about cells, it is not very clear for me.
 A: Given an $m$-dimensional CW-complex $X$, one can lift the CW-structure to a CW-structure on $Y$ by lifting the characteristic maps $\varphi_\alpha : D^k \to X$ to the cover $p : Y \to X$, which can be done since $\pi_1(D^k) \cong 0$.
If degree of $p : Y \to X$ is $n$, there are exactly $n$ lifts of $\varphi_\alpha$ to $Y$. So for each $k$-cell $e^k$ in $X$, there exists $n$ $k$-cells in the lifted CW-structure on $Y$ which are mapped homeomorphically onto $e^k$. 
Let $C_i$ be the number of $i$-cells in $Y$, and $C'_i$ be the number of $i$-cells in $X$. From the above analysis, we derive that $C_i = n \cdot C_i'$ for all $0 \leq i \leq m$. Using the fact that Euler characteristic of a CW-complex $X$, namely the alternating sum of it's betti numbers, is the same as alternating sum of it's number of cells (dimension of the cellular cochain groups), we conclude 
$$\chi(Y) = \sum_{i = 0}^m (-1)^i C_i = \sum_{i = 0}^m (-1)^i n C'_i = n \chi(X)$$
as desired $\blacksquare$

I missed the essential question of OP up there. If $p : Y \to X$ is a covering map, $A \subset X$ is a subspace then $p|_{p^{-1}(A)} : p^{-1}(A) \to A$ is also a covering map. 
In particular, take $A = e^k$ where $e^k$ is one of the cells in $X$. As the only covering space of disks are trivial (of the form $e^k \times D$ where $D$ is a discrete set), and $p$ is of degree $n$, $e^k$ must lift to $n$-many $k$-cells in $Y$.
