[proofplan] We show that the CW structure on $X$ lifts along the $n$-sheeted covering $p: \tilde{X} \to X$ to give a CW structure on $\tilde{X}$ in which each $k$-cell of $X$ has exactly $n$ preimage cells of the same dimension. The alternating sum of cell counts then scales by $n$. [/proofplan] [step:Lift the CW structure of $X$ to $\tilde{X}$ with $n$ copies of each cell] Since $p: \tilde{X} \to X$ is an $n$-sheeted covering and $X$ is a CW complex, the space $\tilde{X}$ inherits a CW structure as follows. For each $k$-cell $e_\alpha$ of $X$ with characteristic map $\Phi_\alpha: D^k \to X$, the covering map provides exactly $n$ lifts $\tilde{\Phi}_{\alpha,1}, \ldots, \tilde{\Phi}_{\alpha,n}: D^k \to \tilde{X}$. Each lift $\tilde{\Phi}_{\alpha,j}$ is a characteristic map for a $k$-cell $\tilde{e}_{\alpha,j}$ of $\tilde{X}$. This is valid because $D^k$ is simply connected (for $k \geq 1$) and contractible, so any map from $D^k$ that starts at a point in a given sheet lifts uniquely. For $k = 0$, each $0$-cell (a point) has exactly $n$ preimages since $p$ is $n$-sheeted. [guided] The key property of covering spaces that we use is the lifting criterion: since $D^k$ is simply connected for $k \geq 1$, the characteristic map $\Phi_\alpha: D^k \to X$ lifts to $\tilde{X}$ once we specify the image of a single point in the fiber. There are exactly $n$ choices for this basepoint (one in each sheet), giving $n$ distinct lifts. Why do these lifts provide a valid CW structure on $\tilde{X}$? The covering map $p$ is a local homeomorphism, so each lift $\tilde{\Phi}_{\alpha,j}$ is a homeomorphism from the interior of $D^k$ onto an open $k$-cell $\tilde{e}_{\alpha,j}$ of $\tilde{X}$, and the restriction to $\partial D^k$ maps into the $(k-1)$-skeleton $\tilde{X}^{k-1}$ (since the attaching map of $e_\alpha$ maps into $X^{k-1}$, and the lift respects skeleta). The cells $\tilde{e}_{\alpha,j}$ are pairwise disjoint (they lie in different sheets over the interior of $e_\alpha$), and they exhaust $\tilde{X}$ because every point of $\tilde{X}$ lies over some cell of $X$. For $k = 0$, the $0$-cells of $X$ are points, and each has exactly $n$ preimages under the $n$-sheeted covering. [/guided] [/step] [step:Compute the Euler characteristic by counting lifted cells] Let $c_k$ denote the number of $k$-cells of $X$. By the previous step, $\tilde{X}$ has $n \cdot c_k$ cells of dimension $k$ for each $k \geq 0$. Therefore \begin{align*} \chi(\tilde{X}) = \sum_{k \geq 0} (-1)^k \cdot n \cdot c_k = n \sum_{k \geq 0} (-1)^k c_k = n \cdot \chi(X). \end{align*} [/step]