I'm working on this problem, from a past Qual.
Let $p:\tilde X\to X$ be a double cover. Show that there is a LES $$\cdots \to H^k(X;\mathbb{Z}/2\mathbb{Z})\to H^k(\tilde X;\mathbb{Z}/2\mathbb{Z})\to H^k(X;\mathbb{Z}/2\mathbb{Z})\to H^{k+1}(X;\mathbb{Z}/2\mathbb{Z})\to\cdots$$ From this deduce that $H^i(\mathbb{R} P^n;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}$ for $i=0,1,\cdots,n$ and is 0 otherwise.
I'm stuck at both parts.
1) I learned about covering maps and cohomology, but I don't recall there is a connection between them. How do I start the first part? I am reading Hatcher.
2) The second part seems wrong. So we know the double cover of $\mathbb{R} P^n$ is $S^n$, and $H^0(S^n,\mathbb{Z}/2\mathbb{Z})=H^n(S^n,\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$ and $0$ elsewhere, which can be computed by Universal COefficients theorem. So we have the exact sequences $$ 0\to H^0(\mathbb{R} P^n,\mathbb{Z}/2\mathbb{Z})\to \mathbb{Z}/2\mathbb{Z}\to H^0(...)\to H^1(...)\to 0 \to... $$ $$ 0\to H^{n-1}(...)\to H^n(...)\to \mathbb{Z}/2\mathbb{Z} \to H^n(...)\to H^{n+1}(...)\to 0 \to ... $$ Here $H^0$ can either be $\mathbb{Z}/2\mathbb{Z}$ or $0$. In the first case, the first map is an isomorphism, hence $H^0$ after that must be $0$, contradiction. In the second case, it would contradicts what we want. Am I wrong somewhre here? This is supposed to be the easy part.