Request for companion of Mariano Suárez-Alvarez's proof. Mariano Suárez-Alvarez's answer to Cohomology of projective plane seems very interesting. However, there are three pieces I could not stitch up for one of his proofs. Wonder if someone may help?

Later: alternatively, one can do a bit of magic. Since there is a covering $S^2\to P^2$ with $2$ sheets, we know that the Euler characteristics of $S^2$ and $P^2$ are related by $\chi(S^2)=2\chi(P^2)$. Since $\chi(S^2)=2$, we conclude that $\chi(P^2)=1$. Since $P^2$ is of dimension $2$, we have $\dim H^p(P^2)=0$ if $p>2$; since $P^2$ is non-orientable, $H^2(P^2)=0$; finally, since $P^2$ is connected, $H^0(P^2)\cong\mathbb R$. It follows that $1=\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)=1-\dim H^1(P^2)$, so that $H^1(P^2)=0$.

So my questions are:
Why

since $P^2$ is non-orientable, $H^2(P^2)=0$
since $P^2$ is connected, $H^0(P^2)\cong\mathbb R$
$\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)$?

Thank you very much!
 A: *

*[What follows doesn't actually address the question, see below] An orientation on a $n$-dimensional manifold is given by a nowhere zero differential $n$-form; such a form is called a volume form and is necessarily closed. If the manifold is compact, as $P^2$ is, such a form cannot be exact, otherwise the manifold would have zero volume by Stokes' Theorem which can't happen unless $n = 0$. As such, on a compact manifold $X$, a volume form defines a non-trivial cohomology class in $H^n_{\text{dR}}(X)$, so a compact, non-orientable manifold must have $H^n_{\text{dR}}(X) = 0$.

*For a manifold $X$, $H^0_{\text{dR}}(X) = \operatorname{ker} d: \Omega^0(X) \to \Omega^1(X)$. That is, $H^0_{\text{dR}}(X) = \{f \in C^{\infty}(X) \mid df = 0\}$. Note that $df = 0$ if and only if $f$ is locally constant. If $X$ is connected, a locally constant function is constant so $$H^0_{\text{dR}} = \{f : X \to \mathbb{R} \mid f(X) = \{a\}\ \text{for some}\ a \in \mathbb{R}\} \cong \mathbb{R}$$ where the isomorphism takes a constant function and sends it to its value. More generally, if $X$ has $k$ components, $H^0_{dR}(X) \cong \mathbb{R}^k$. 

*The formula for the Euler characteristic of an $n$-dimensional manifold $X$ is $\chi(X) = \sum_{i=0}^n(-1)^i\operatorname{dim}H^i_{\text{dR}}(X)$. If $n = 2$ (i.e. $X$ is a surface), this reduces to $$\chi(X) = \operatorname{dim}H^0_{\text{dR}}(X) - \operatorname{dim}H^1_{\text{dR}}(X) + \operatorname{dim}H^2_{\text{dR}}(X).$$ For $X = P^2$ we have $\operatorname{dim}H^2_{\text{dR}}(P^2) = 0$ from point one which gives $$\chi(P^2) = \operatorname{dim}H^0_{\text{dR}}(P^2) - \operatorname{dim}H^1_{\text{dR}}(P^2)$$ and from point two we have $\operatorname{dim}H^0_{\text{dR}}(P^2) = 1$ so $$\chi(P^2) = 1 - \operatorname{dim}H^1_{\text{dR}}(P^2).$$

Added Later: As ronno correctly pointed out, my argument in point 1 is for the converse. That is, I explain that an orientable manifold must have non-trivial top degree cohomology because a volume form defines a non-trivial cohomology class. However, that does not imply that the top degree cohomology of a non-orientable manifold must be zero as it could a priori have non-trivial cohomology classes which are not representable by volume forms. 
The desired result is actually a bit harder and requires the orientable double cover of a non-orientable manifold. See Theorem $17.34$ of Lee's excellent book Introduction to Smooth Manifolds (second edition) for a detailed proof.
