Rational cohomology of punctured closed non-orientable manifold. Let $M-\{p\}$ be a punctured closed non-orientable even-dimensional manifold. If $M=\mathbb{R}{P}^{2}$ then $\mathbb{R}P^{2}-\{p\}$ is an open Mobius strip. This implies that $H^{*}(\mathbb{R}P^{2}-\{p\};\mathbb{Q})\cong H^{*}(S^{1};\mathbb{Q}).$  If the cohomology of $M$ is given then how can we relate the rational cohomology of $M-\{p\}$ with $M$? For example, if we take $M=\mathbb{R}P^{2}\times M_{g}.$ Here, $M_{g}$ is closed orientable surface.
I have the following information about the rational cohomology. Maybe my information is wrong.
Let $dim(M)=n$. Its obvious that $H^{n}(M-{\{p\}};\mathbb{Q})$ is zero. The map $H^{i}(M;\mathbb{Q})\rightarrow H^{i}(M-{\{p\}};\mathbb{Q})$ is an isomorphism for $i<n-1.$ I want to know about the cohomology group $H^{n-1}(M-{\{p\}};\mathbb{Q}).$ In the case of real projective plane, this is isomorphic to $\mathbb{Q}$. Is it true that $H^{n-1}(M-{\{p\}};\mathbb{Q})\neq0$ for any closed non-orientable manifold $M?$ For example, $H^{3}(\mathbb{R}P^{2}\times M_{g}-\{p\};\mathbb{Q})\neq0?$
 A: Yes, $H^{n-1}(M-\{p\};\mathbb{Q}) \neq 0$ for any closed non-orientable manifold of dimension $n$. Long-story-short it's because $H^n(M)=0$ in this case so $H^{n-1}(M-\{p\})$ surjects onto the coefficient group. 
Just for exposition I wanted to spell out the arguments to a few of your assertions. Consider the long exact sequence of the pair:
$$\dots \to H^{k}(M, M-\{p\}) \to H^k(M) \to H^k(M-\{ p\})$$$$ \to H^{k+1}(M, M-\{p\}) \to H^{k+1}(M)\to \dots $$
This sequence exists for all coefficients $R$, in particular $\mathbb{Q}$.
Since $M$ is a manifold of dimension $n$, the group $H^k(M, M-\{p\})$ is $R$ if $k = n$ and $0$ otherwise. It follows that 
$$H^k(M) \cong H^k(M-\{p\})\text{ if $k < n-1$}$$
and there is an exact sequence 
$$0 \to H^{n-1}(M) \to H^{n-1}(M-\{ p\}) \to R \to H^n(M) \to 0$$
where $H^n(M-\{p\}) = 0$ because it is an open manifold. 
If $M$ is orientable then the last map is an isomorphism so $H^{n-1}(M) \cong H^{n-1}(M-\{p\})$. Otherwise $M$ is non-orientable, so if $R$ is such that $H^n(M) =0$ (for example $R=\mathbb{Z}, \mathbb{Q}$) then you get a short exact sequence with $H^{n-1}(M-\{p\})$ in the middle: if $R=\mathbb{Q}$ that means $H^{n-1}(M-\{p\})$ is a rational vector space of dimension $dim(H^{n-1}(M)) +1>0$. In fact as long as $H^n(M)=0$ and $R\neq 0$ we must have $H^{n-1}(M-\{p\};R)\neq 0$ because it surjects onto $R$.
