An injection yields an isomorphism in cohomology? $\Bbb RP^n \rightarrow \Bbb RP^{\infty}$.

Question

In page 79 line 10 Hatcher made the claim that

An embedding $\Bbb RP^{n-1} \rightarrow \Bbb RP^\infty$ induces an isomoprhism $$H^i(\Bbb RP^n, \Bbb Z_2) \leftarrow H^i(\Bbb RP^\infty, \Bbb Z_2)$$ for $i\le n$.

I know both cohomology groups are isomorphic to $\Bbb Z_2$. I also know this holds when we have the canonical cell inclusion from cellular cohomology. But I don't understand how this is true for arbitrary embedding.

Answer 1

If you take any open contractible subset $U$ of a manifold $M$, the injection $U\rightarrow M$ does not necessarily induces an isomorphism on any cohomology group. The CW-structure on $\mathbb{R}P^{\infty}$ is the limit of the CW-structure on $\mathbb{R}P^n$. The injection $\mathbb{R}P^n\rightarrow \mathbb{R}P^{\infty}$ identifies the $i$-cell of $\mathbb{R}P^n$ with the $i$-cell of $\mathbb{R}P^{\infty}$, $i<n+1$. this enables to show that result.