TITLE: Homology of real projective space... I'm not satisfied with the argument in hatcher.
QUESTION [21 upvotes]: In example 2.42 Hatcher computes the homology of real projective space. I follow his argument, but I would be uncomfortable believing the details of the degree computation if I didn't see it in his text.
How does Hatcher conclude that the degree of one of the local homeomorphisms is 1, as opposed to -1? (I realize that this doesn't actually affect the homology computation.) 
Is there some topological fact that lets him conclude that the restriction to one hemisphere of the composition $S^{k-1} \to RP^{k-1} \to RP^{-1}/RP^{k-2} = S^{k-1}$ is actually the identity map?

REPLY [2 votes]: Here is an explicit computation. For $k\geq 2$,  let $\varphi_k:S^{k-1}\to\mathbb{RP}^{k-1}$ denote the attaching map for the $k$-cell of $\mathbb{RP}^n$:
 $$\varphi_k(x_1,\cdots,x_k)=[x_1,\cdots,x_k].$$ 
The cellular boundary formula says that the degree of the boundary homomorphism
$d_k:H_k(\mathbb{RP}^k,\mathbb{RP}^k-1)\to H_{k-1}(\mathbb{RP}^{k-1},\mathbb{RP}^k-2)$  is  equal (upto sign, which is irrelevant to the compuation of homology groups) to the degree of the composite
 $$S^{k-1}\xrightarrow{\varphi_{k}}\mathbb{RP}^{k-1}\xrightarrow{q}\mathbb{RP}^{k-1}/\mathbb{RP}^{k-2},$$
where $q$ denotes the quotient map which collapses $\mathbb{RP}^{k-2}$ to a point and $M:=\mathbb{RP}^{k-1}/\mathbb{RP}^{k-2}$ is endowed with some orientation. Since $(q\circ \varphi_k)^{-1}(q([0,\cdots,0,1]))={N,S}$, where $N=-S=(0,\cdots,0,1)$ are the north and the south poles of $S^{k-1}$, the degree is equal to  the sum of the local degrees at $N$ and $S$: $\deg(q\circ\varphi_k)=\deg(q\circ \varphi_k)\vert_N+\deg(q\circ \varphi_k)\vert_S$.
To compute the local degrees, observe that  $\varphi_k\circ\alpha=\varphi_k$, where $\alpha:S^{k-1}\to S^{k-1}$ is the antipodal map. Since $\deg\alpha=(-1)^k$,  we have $$\deg(q\circ \varphi_k)\vert_S=\deg(q\circ \varphi_k\circ\alpha)\vert_S=\deg(q\circ \varphi_k)\vert_N\deg(\alpha)\vert_S=(-1)^k\deg(q\circ\varphi_k)\vert_N.$$ Thus what remains is to compute the local degree at $N$. Let $\ast=q(\mathbb{RP}^{k-2}).$ We can identify (i.e. there is a homeomorphism) $M\setminus{\ast}$ with $\mathbb{R}^{k-1}$ via $q([y_1,\cdots,y_{k-1},1])\mapsto(y_1,\cdots,y_{k-1})$. We can also identify the upper hemisphere of $S^{k-1}$ with the open unit ball $B^{k-1}$ via the projection $(x_1,\cdots,x_{k})\mapsto(x_1,\cdots,x_{k-1})$. Under these identification, the restriction of $q\circ\varphi_{k}$ to the upper hemisphere is given by $$B^{k-1}\to\mathbb{R}^{k-1},\, (x_1,\cdots,x_{k-1})\mapsto(\frac{x_1}{\sqrt{1-\lVert x\rVert^2}},\cdots,\frac{x_{k-1}}{\sqrt{1-\lVert x\rVert^2}}).$$ This is a homeomorphism, so its local degree at any point is $1$ (upto sign).