Cohomology of Real Projective Space

This is from Hacther Theorem 3.19 (Page 220), a calculation of the cohomology ring $P^n$ (abbreivation for $\mathbb{R}P^n$) with $\mathbb{Z}_2$ coefficients. I have two quick questions:

  1. He first writes that "the inclusion $P^{n-1}\hookrightarrow P^n$ induces an isomorphism on $H^i$ for $i\leq n-1$." I can understand how this works since I know the explicit cohomology groups to be $\mathbb{Z}_2$. But is there a more general theorem that applies when relating subcomplexes of CW complexes? From celluluar homology it seems that we can get it for $i<n-1$ for the inclusion $X^i\hookrightarrow X$, as the $(i+1)$-cells have bearing on $H^i$.
  2. He then writes that "it suffices by induction on $n$ to show that the cup product of a generator of $H^{n-1}(P^n)$ with a generator of $H^1(P^n)$ is a generator of $H^n(P^n)$. But how does this take into account $H^{n-2}(P^n)$ and $H^2(P^n)$, for example? I think I'm confused by what "induction on $n$" really is, since there's $H^n$ and $P^n$.

  1. Yes, for $i<n-1$, there are isomorphisms $H^i(X) \to H^i(X^n)$, since cellular cohomology defines $H^i$ in terms of data from only the $(i+1)$-skeleton. What is going on for $P^n$, though, is that the degrees of the attachment maps are all $2$, so the coboundary maps are all trivial. That means that the $i$th cohomology of $P^n$ is determined by its $i$-skeleton.

  2. If you have that the generator for $H^n(P^n)$ is obtained by taking the cup product of the $H_1(P^n)$ generator with itself, then you have the same property for the generator for $H^n(P^{n+k})$ for all $k$ due to the above isomorphisms. That means we just need to show the property holds for the generator of $H^n(P^n)$ for all $n$. For example, $H^{n-2}(P^n)$ would be handled by $H^{n-2}(P^{n-2})$ and $H^2(P^n)$ would be handled by $H^2(P^2)$. Or, if we're taking the cup product of the $H^2(P^n)$ generator with the $H^{n-2}(P^n)$ generator, then we can reassociate it because by hypothesis they're both repeated cup products of the $H^1(P^n)$ generator.