Calculate the map on homology induced by the quotient map $S^
3 → \Bbb R {\Bbb P}^3$
I understand the quotient map $q:S^3 → \Bbb R {\Bbb P}^3$ i.e. identifying antipodal points i.e. for $x \in S^3$ consider te corresponding $[x] \in \Bbb R {\Bbb P}^3$ then, $q(x)=q(-x)=[x] , \forall x \in S^3$ .
It is a 2-sheeted covering map and hence should give a degree 2 map and hence in the level of homology, it should behave like $n \mapsto 2n$ .But not quite clear to me.
I have visited this question but it has a few comments, I want to understand this concept and so if someone vance for help!can write a detailed answer about what is going on here and how to compute it, it will gteatly help my cause ( also to understand similar problems) . Thanks in advance for help!
Best Answer
In general the canonical quotient map $q:S^n\rightarrow\mathbb{R}P^n$ is the attaching map for the top cell of $\mathbb{R}P^{n+1}$. This means that there is a cofibration sequence
$$S^n\xrightarrow{q}\mathbb{R}P^n\rightarrow \mathbb{R}P^{n+1}$$
and in particular a long exact sequence in homology
$$\dots\rightarrow H_{k+1}\mathbb{R}P^{n+1}\rightarrow H_kS^n\xrightarrow{q_*}H_k\mathbb{R}P^n \rightarrow H_k\mathbb{R}P^{n+1}\rightarrow H_{k-1}S^n\rightarrow\dots$$
If we put a CW structure on $S^n$ with two cells in each dimension $0\leq r\leq n$ then these cells are permuted by the $\mathbb{Z}_2$-action of the antipodal map. Hence the structure descends to give a CW structure of the quotient $\mathbb{R}P^n$ which has one cell in each dimension $0\leq r\leq n$.
Then in $S^n$, the boundary orientation of the two $r$-cells agrees if $r$ is even, and is opposite in the case that $r$ is odd. This means that the cellular complex $C_*S^n$ (whose homology computes $H_*S^n$) is actually exact in degrees $\neq 0,n$. On the other hand, it means that in $C_*\mathbb{R}P^m$, where the cells in each dimension are identified, the differentials are either trivial (when the orientations are opposite), or multiplication by $2$ (when the orientations agree). Hence we compute
$$\tilde H_k\mathbb{R}P^{2r}\cong\begin{cases}0&k\leq 2r\:\text{even}\\\mathbb{Z}_2&k\leq 2r\:\text{odd}\end{cases}$$
and
$$\tilde H_k\mathbb{R}P^{2r+1}\cong\begin{cases}0&k< 2r+1\:\text{even}\\\mathbb{Z}_2&k<2r+1\;\text{odd}\\ \mathbb{Z}&k=2r+1\end{cases}$$
This gives us that whenever $n=2r+1$ is odd we have an exact sequence
$$H_{2r+2}\mathbb{R}P^{2r+2}\rightarrow H_{2r+1}S^{2r+1}\xrightarrow{q_*} H_{2r+1}\mathbb{R}P^{2r+1}3\rightarrow H_{2r+1}\mathbb{R}P^{2r+2}\rightarrow H_{2r}S^{2r+1}$$
which is exactly
$$0\rightarrow \mathbb{Z}\xrightarrow{q_*}\mathbb{Z}\rightarrow \mathbb{Z}_2\rightarrow 0.$$
Hence $q_*:H_{2r+1}S^{2r+1}\rightarrow H_{2r+1}\mathbb{R}P^{2r+1}$ is multiplication by $2$. In particular this holds for the odd integer $n=3$.