What does it mean that the quotient $S^n\to\mathbb{R}P^n$ acts as the identity on the upper hemisphere, and the antipodal map on the lower hemisphere?

Question

I'm not sure how the degree of cellular maps are computed when finding the homology of $\mathbb{R}P^n$.

I know $RP^n$ has CW structure with a cell in each degree, and $e^k$ is glued to $RP^{k-1}$ by the attaching map $q:S^{k-1}\to RP^{k-1}$.

The degree of the $k$th cellular map $d_k$ is the degree of the composition $$ S^{k-1}\stackrel{q}{\to}RP^{k-1}\to RP^{k-1}/RP^{k-2}=S^{k-1} $$

Every explanation I come across concludes that the degree is $\deg(i)+\deg(a)=1+(-1)^{k-1}$, where $i$ and $a$ are the identity and the antipodal map.

The reasoning is that the composition acts as the identity on the upper hemisphere, but as the antipodal map on the lower hemisphere? What does that mean? If $x\in S^{k-1}$, then $$ x\mapsto [x]=\{x,-x\}\mapsto\ ? $$ I don't know what the second map does on the equivalence classes to make the composition act like the identity and antipodal map as described above. I suppose it just sends $[x]\mapsto x$ if $x$ is in the upper hemisphere? How do we know that's what the second quotient map does?

Thanks.

Answer 1

The crucial point is how $RP^{k-1}/RP^{k-2}$ is identified with $S^{k-1}$.

$RP^{k-1}$ is the quotient of the northern hemisphere $N$ that is obtained by identifying opposite points on its boundary $S^{k-2}$. If $RP^{k-2}$ is divided out, this is the same as building the quotient $N/S^{k-2}$, which is the same as $S^{k-1}$ since it is the one point compactification of an open disk.

Using this identification, $N$ is mapped onto $S^{k-1}$ by "stretching" (so you're map isn't really the identity on the hemisphere, just homotopic to it) it towards the south pole, and $S$ is mapped onto $S^{k-1}$ by first applying $-id$ and then doing the same.