We write $P^k$ to denote $\mathbf RP^k$ and always work with $\mathbf Z/2\mathbf Z$ coefficients.

On pg 221 of Hatcher'sAlgebraic Topology, there is the following diagram

Here by $P^{i-1}$ we mean the subspace of $P^n$ consisting of points represented by vectors in $\mathbf R^{n+1}$ whose last $n-(i-1)$ coordinates are $0$. Similarly for $P^i$.By $P^j$ we mean the subspace of $P^n$ consisting of points represented by vectors in $\mathbf R^{n+1}$ having the first $i$ coordinates $0$.

Question.Hatcher says that the maps in the left-hand square are all isomorphismsbecause of cellular cohomology.

I am stuggling to see how. The way I can see the top map in the left hand square is an isomorphism is via the following argument. We have part of the long exact sequence for the pair $(P^n, P^{i-1})$

$$H^i(P^n, P^{i-1})\rightarrow H^i(P^n)\rightarrow H^i(P^{i-1})\rightarrow \cdots$$

The group $H^i(P^n, P^{i-1})$ is $\mathbf Z/2\mathbf Z$ because $H^i(P^n, P^{i-1})\cong H^i(P^n/P^{i-1}))$. Now $P^n/P^{i-1}$ is homeomorphic to the wedge sum of shperes, one for each of the dimensions $i, i+1, \ldots, n$. The group $H^i(P^n)$ is $\mathbf Z/2\mathbf Z$ because of the universal coefficient theorem and because $H_i(P^n)=\mathbf Z/2\mathbf Z$.Lastly, $H^i(P^{i-1})=0$ because of cellular cohomology and the fact that $P^{i-1}$ has no cell in the $i$-th dimension. Thus the map $H^i(P^n, P^{i-1})\to H^i(P^n)$ is a surjection, and since both groups are $\mathbf Z/2\mathbf Z$, this map is an isomorphism. Similarly one can show that the bottom map in the left-hand square is an isomorphism. Is there an easier way to see that these maps are isomorphisms? Thank you.