I am currently learning about homology theory. I have started with the definition of singular homology, and I am trying to compute the homology of some spaces we often use, but I am stuck with the real projective spaces.
The result I want to get is:
if $n$ is odd, then $H_k(\mathbb{R}\mathbb{P}^n)\cong \left\{\begin{array}{cc} \mathbb{Z} & \textrm{if } k\in\{0,n\} \\ \mathbb{Z}/2\mathbb{Z} &\textrm{if } 0<k<n \textrm{ and } k\textrm{ odd}\\ 0 & \textrm{else } \end{array} \right.$
if $n$ is even, then $H_k(\mathbb{R}\mathbb{P}^n)\cong \left\{\begin{array}{cc} \mathbb{Z} & \textrm{if } k=0 \\ \mathbb{Z}/2\mathbb{Z} & \textrm{if } 0<k<n \textrm{ and } k\textrm{ odd}\\ 0 & \textrm{else } \end{array} \right.$
I did complex ones, and so I started with the same ideas. ISo I define $A=\{[x_0:\ldots:x_n]\vert x_0\ne 0\}$ which is homeomorphic to $\mathbb{R}^n$ and $B=\mathbb{R}\mathbb{P}^n\setminus\{[1:0:\ldots:0]\}$ which has the homotopy type of $\mathbb{R}\mathbb{P}^{n-1}$. Then, $A\cap B=A\setminus\{[1:0:\ldots:0]\}$ has the homotopy type of the sphere $S^{n-1}$. So we want to proceed by induction.
For the case $n=1$, I know that $\mathbb{R}\mathbb{P}^1$ is homeomorphic to the circle so that gives what I want.
For the projective plane, I get a sequence $0 \to H_2(\mathbb{R}\mathbb{P}^n) \to \mathbb{Z} \to 0\oplus \mathbb{Z} \to H_1(\mathbb{R}\mathbb{P}^n) \to \cdots$
The sequence does not determine the homology groups, so I have to understand the map $\mathbb{Z} \to \mathbb{Z}$, mapping a generator of the homology of the circle, to the homology of the projective line. Although the projective line is homeomorphic to the circle, we have $\mathbb{R}\mathbb{P}^1=S^1/\pm1$ so so in fact the circle wraps twice in the projective line, giving the map being multiplication by two and the result follow.
Now for $n>2$: most of the groups are determined by the sequence, except for $H_n$ and $H_{n-1}$, when $n$ is even. There, we get $0 \to H_n(\mathbb{R}\mathbb{P}^n) \to H_{n-1}(S^{n-1})\cong\mathbb{Z} \to 0\oplus H_{n-1}(\mathbb{R}\mathbb{P}^{n-1})\cong \mathbb{Z} \to H_1(\mathbb{R}\mathbb{P}^n) \to 0$
The situation is the same as before, and we must have the map being a multiplication by two to get the result, but I do not see how I can say that. What I saw, following the maps all the way through with inclusions, homeomorphism and homotopy equivalence, that the map is exactly the one induced by the projection $S^{n-1} \to S^{n-1}/\pm 1$. The answer might be the same as before, but while I could picture the circle wrapping twice on the projective line, I do not see what happens in higher dimensions.
So, how does one see that it is indeed a multiplication by 2 ?
Edit : following the comment on degree, thinking about what the identity map and the antipodal map does gave me an idea of how to picture this (which might be the degree ?): instead of seeing $\mathbb{R}\mathbb{P}^{n-1}$ as a quotient of a sphere, it is also $D^{n-1}$ the close disk in $\mathbb{R}^{n-1}$, with the identification done on it boundary $S^{n-2}$ (because the northern hemisphere along with the equator suffices to get any point up to antipodal map). So then, we send the sphere to its northern hemisphere. What we can show is that the $n-1$ homology of the sphere i generated by $a-b$, $a$ being a map to the northern hemisphere, and $b$ to the southern one. Then, each $a$ is sent to some simplex $x$, and $b$ is sent to $\pm x$ - here, $b$ is changed by the antipodal map so this $\pm 1$ would be what is called the degree of the antipodal map ? Anyway, in our case this will be sent to $x$ too. So we send our generator to $2x$, but then, one would need to see that $x$ (being $a$ but since we have the identification on the boundary, it is not exactly the northern hemisphere) is a generator in $\mathbb{R}\mathbb{P}^{n-1}$ ?
