Calculate the map on homology induced by the quotient map $S^ 3 → \Bbb R {\Bbb P}^3$

Question

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!

Answer 1

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 $n=3$.

Answer 2

Along with Tyrone's comprehensive computation, you can also see it at the level of cell-structures since it's just a finite cover. The argument I give here is somewhat along the lines of arguing using the transfer map for finite covers. (All of the chain and homology groups are cellular.)

Choose a CW structure for $S^n$ which has two cells in each dimension $0\leq i \leq n$ and so that the antipodal map $a\colon S^n \to S^n$ is cellular. Then there is an induced cell-structure on $\mathbb{R}P^n$ with one cell in each dimension $0\leq i \leq n$ and so that the quotient map $q\colon S^n \to \mathbb{R}P^n$ is cellular.

Now, $H_k(S^n)$ is only non-zero in two degrees: $0$ and $n$.

In $H_0$, $q$ induces an isomorphism since both spaces are connected.

For $H_n(S^n)$, given the $2$-cell $\varphi_+\colon D^n \to S^n$ representing the upper hemisphere, then the sum $\varphi_+ + a_*(\varphi_+) \in C_n(S^n)$ is a cycle, and generates $H_n$. If $\varphi_P\in C_n(\mathbb{R}P^n)$ is the elementary chain given by the unique $2$-cell, then $$q_*(\varphi_+ + a_*(\varphi_+)) = 2 \varphi_P$$

Since $\varphi_P$ is the only $n$-cell, it and its multiples are the only candidates for elements of $H_n(\mathbb{R}P^n)$, but in fact it is a cycle iff the antipodal map on $S^{n-1}$ reverses orientation (otherwise its boundary is $2$ times the generator of $C_{n-1}(\mathbb{R}P^n)$), iff $n$ is odd. Therefore if $n$ is even then $H_n(q)\colon \mathbb{Z} \to 0$ is the $0$ map, and if $n$ is odd then $[\varphi_P]$ generates $H_n(\mathbb{R}P^n)\cong \mathbb{Z}$ and $H_n(q)\colon \mathbb{Z} \to \mathbb{Z}$ is multiplication by $2$.