I was trying to compute the homology groups of $P^n$ by induction using the long exact sequence of homology (Without using cellular homology, although some cellular reasoning can be used, if it can be justified in a simple way without cellular homology). I want to prove that $H_m(P^n)=\mathbb{Z_2}$ for $m<n$ odd, $\mathbb{Z}$ for $m=0$, $0$ for $m$ even and $\mathbb{Z}$ if $m=n$ is odd. The case $n=2$ has already been proven and $P^1\cong S^1$. My attempt goes like this:
Let first $n>1$ be odd, and let's suppose for every $k<n$ the homology groups are as stated above. Then, using the fact that $P^{n-1}\subseteq P^n$ (As the projection of the points in $S^n$ which have a fixed coordinate null) and that $P^n/P^{n-1}\cong S^n$ we have a long exact sequence of (reduced) homology: $$\cdots \to H_m(P^{n-1})\to H_m(P^n)\to H_m(S^n)\to H_{m-1}(P^{n-1})\to \cdots$$ and in particular, taking $m=n$ we have $$\cdots \to H_n(P^{n-1})\to H_n(P^n)\to H_n(S^n)\to H_{n-1}(P^{n-1})\to \cdots$$ which gives us ($n-1$ is even), since $H_n(P^{n-1})=H_{n-1}(P^{n-1})=0$, the exact sequence $$0\to H_n(P^n)\to \mathbb{Z}\to 0$$ and this gives an isomorphism between $\mathbb{Z}$ and $H_n(P^n)$.
Similarly, for $1\leq m<n$ with $m$ odd (If this case is possible), and since $H_m(P^{n-1})=\mathbb{Z}_2$ we have the exact sequence $$0 \to \mathbb{Z}_2\to H_m(P^n)\to 0$$ and this gives an isomorphism between $\mathbb{Z}$ and $H_n(P^n)$.
Lastly, for $1<m<n$ even, and since $H_m(P^{n-1})=H_m(S^n)=0$ we have the exact sequence $$0 \to H_m(P^n)\to 0$$ which means that $H_m(P^n)=0$.
So, for $n$ odd the job is done.
For $n$ even, I try to do it similarly, but there is a problem.
Similarly, take $n$ even and suppose that for every $k<n$ the homology groups are as stated above. Then take the long exact sequence: $$\cdots \to H_m(P^{n-1})\to H_m(P^n)\to H_m(S^n)\to H_{m-1}(P^{n-1})\to \cdots$$ For $1\leq m< n-1$ odd we have $H_m(P^{n-1})=\mathbb{Z}_2$ and $H_m(S^n)=0$. Also $H_{m+1}(S^{n})=0$. So we have the exact sequence $$0 \to H_m(P^{n-1})\to H_m(P^n)\to 0$$ which gives an isomorphism between $H_m(P^{n-1})$ and $H_m(P^n)$, so both of them are $\mathbb{Z}_2$. For $1<m<n$ even we have, since $H_m(P^{n-1})=0$ and $H_m(S^n)=0$, the sequence $$0\to H_m(P^n)\to 0$$ which results in $H_m(P^n)=0$.
The homology groups we haven't calculated yet are just the ones at the start of the long exact sequence, which is, since $H_n(P^{n-1})=0$, $$0\to H_n(P^n)\to \mathbb{Z}\stackrel{\partial}{\to} \mathbb{Z}\to H_{n-1}(P^n)\to 0.$$
So I want to prove the homomorphism $\partial$ is just multiplication by $2$. I can also see this as $\partial:H_n(P^n,P^{n-1})\to H_n(P^{n-1})$ and as $\partial: H_n(S^n)\to H_n(P^{n-1})$, then I have to pick a generator of either domain and see its image. But I'm not sure how can I do that.
The connecting map $\delta$ lifts a cycle in $S^n$ to a preimage in $P^n$, then takes its boundary, as an element of $P^{n-1}$.
So for $\delta: H_n(S^n)\rightarrow H_n(P^{n-1})$, you lift the generating class $[c]$ of $H_n(S^n)$ itself to the ball with boundary that is the closure of $P^n\setminus P^{n-1}$. Thus $\delta([c])$ is then $d\cdot [b]$, where $[b]$ generates $H_{n-1}(P^{n-1})$, and $d$ is the degree of the covering $S^{n-1}\rightarrow P^{n-1}$, treating $S^{n-1}$ as the boundary of this ball pre-quotient.