In Hatcher p.221, with the notation $\mathbf{P}^n$ for $\mathbb{RP}^n$, and all cohomology groups in $\mathbb{Z}_2$, he says
I can see from the l.e.s. on the cohomology of the pair $(\mathbf{P}^n, \mathbf{P}^{n-1})$ the exact sequence
$0 = H^n(\mathbf{P}^{n-1})\leftarrow H^n(\mathbf{P}^n)\xleftarrow{j^*} H^n(\mathbf{P}^n, \mathbf{P}^{n-1})\xleftarrow{\partial} H^{n-1}(\mathbf{P}^{n-1})\xleftarrow{i^*} H^{n-1}(\mathbf{P}^n)$
But then I would need to have $\partial = 0$ for the map $j^*$ to be an isomorphism.
This could be established by showing that the map $i^*$ is an isomorphism.
I am assuming this is what Hatcher means, also based on the answer in A Step in Calculating the Cup Product on $\mathbf RP^n$.
How do we know $i^*$ is indeed an isomorphism? As pointed out, this is mentioned in the second line of Hatcher's proof:
"The inclusion $\mathbf{P}^{n-1}\to \mathbf{P}^n$ induces an isomorphism on $H^i$, for $i\leq n-1$."
Would someone clarify why this statement is true?