Homology of real projective space... I'm not satisfied with the argument in hatcher.

Question

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?

Answer 1

caveat: I am not that comfortable with this answer.

I think something is hidden in the notation here. The identification $$ \mathbb{RP}^{k-1}/\mathbb{RP}^{k-2} = S^{k-1} $$ is not canonical, despite what is suggested by the equals sign. For a concrete example, consider $\mathbb{RP}^2/\mathbb{RP}^1$.

If we take a point on $\mathbb{RP}^2$, thought of as a pair of antipodal points on $S^2$, and we map it to its representative on the southern hemisphere (well-defined unless it's on the equator, which we're killing),then collapse the closure of the northern-hemisphere to a point, we get a different identification of $\mathbb{RP}^2$ with the sphere than if we do the same with the words "northern" and "southern" replaced everwhere. These two maps differ by precomposition by the antipodal map. So one must make an arbitrary choice of identification (note that up to homotopy, this only matters for half of the $k$ (I am confused about odd and even, doubly so because our index starts on $k - 1$)).

Answer 2

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).

Answer 3

Here is a way to compute the cellular chain complex that is careful with signs. I'm not sure how useful it is, but perhaps it is nice to know that it is possible.

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I first want to address the issue raised by hunter's answer. Actually, there is a way to ensure that the identification $\mathbb{R}P^n/\mathbb{R}P^{n - 1} \cong S^n$ is canonical. This requires that we fix a homeomorphism $D^n / S^{n - 1} \cong S^n$, but this is something we should always do anyway when working with cellular homology. (See the treatment in May's A Concise Course in Algebraic Topology, which is very careful about this.)

After fixing the homeomorphism $D^n / S^{n - 1} \cong S^n$, we use the fact that there is a fixed characteristic map $\Phi^n : (D^n, S^{n - 1}) \to (\mathbb{R}P^n, \mathbb{R}P^{n - 1})$ associated to the $n$-cell of $\mathbb{R}P^n$. Taking quotients on each side, $\Phi^n$ descends to a homeomorphism $D^n / S^{n - 1} \cong \mathbb{R}P^n / \mathbb{R}P^{n - 1}$. By composing with our fixed homeomorphism $D^n / S^{n - 1} \cong S^n$, we obtain the homeomorphism $S^n \cong \mathbb{R}P^n / \mathbb{R}P^{n - 1}$.

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With this established, we can compute the cellular chain complex of $\mathbb{R}P^n$ as follows. We will need two key models of $S^n$, which are $D^n / S^{n - 1}$ and $D^n \cup_{S^{n - 1}} D^n$. Thus, we should fix homeomorphisms $D^n / S^{n - 1} \cong S^n$ and $D^n \cup_{S^{n - 1}} D^n \cong S^n$. Let us use the homeomorphisms depicted in the following picture:

In words: The homeomorphism $D^n / S^{n - 1} \cong S^n$ sends the ray from $0$ to $x$ to the longitude running from north to south that passes through the point $(x, 0)$. The homeomorphism $D^n \cup_{S^{n - 1}} D^n \cong S^n$ sends the first copy of $D^n$ to the northern hemisphere of $S^n$ and the second copy of $D^n$ to the southern hemisphere, in the obvious way.

We also need to fix the CW structure for $\mathbb{R}P^n$. This is done inductively, by defining the characteristic map $\Phi^k : (D^k, S^{k - 1}) \to (\mathbb{R}P^k, \mathbb{R}P^{k - 1})$ associated to the single $k$-cell as the composite $$ D^k \hookrightarrow D^k \cup_{S^{k - 1}} D^k \cong S^k \to \mathbb{R}P^k, $$ where the first map is the inclusion of the first factor, the second map is the homeomorphism fixed above, and the third map is the quotient map. In other words, $\Phi^k$ is the restriction of the quotient map $S^k \to \mathbb{R}P^k$ to the northern hemisphere.

Now, to compute the cellular boundary map, we want to compute the degree of the map $$ f_k : S^{k - 1} \xrightarrow{\varphi^k} \mathbb{R}P^{k - 1} \to \mathbb{R}P^{k - 1} / \mathbb{R}P^{k - 2} \xleftarrow[\cong]{\Phi^{k - 1}} D^{k - 1} / S^{k - 2} \cong S^{k - 1}, $$ where $\varphi^k$ is the attaching map and $\Phi^{k - 1}$ is the characteristic map. As suggested by Hatcher, let us restrict this map to the northern hemisphere of $S^{k - 1}$, which by our second model is identified with $D^{k - 1}$: $$ f_k|_{D^{k - 1}} : D^{k - 1} \xrightarrow{\varphi^k} \mathbb{R}P^{k - 1} \to \mathbb{R}P^{k - 1} / \mathbb{R}P^{k - 2} \xleftarrow[\cong]{\Phi^{k - 1}} D^{k - 1} / S^{k - 2} \cong S^{k - 1}. $$ Observe that the first map is just $\Phi^{k - 1}$! Thus, this composition is just the quotient map $D^{k - 1} \to S^{k - 1}$ we fixed at the beginning.

We want to compute the local degree of $f_k$ at the north pole $N$. Let $U$ be the interior of the northern hemisphere of $S^{k - 1}$, and let $V$ be $S^{k - 1}$ minus the south pole. Then the local degree is given by the map $$ H_{k - 1}(U, U \setminus \{N\}) \xrightarrow{(f_k|_U)_*} H_{k - 1}(V, V \setminus \{N\}). $$ From the analysis in the previous paragraph, we know that $f_k$ restricts to the quotient map $D^{k - 1} \to S^{k - 1}$, which further restricts to a homeomorphism $U \cong V$. In particular, $f_k|_U$ is homotopic to the inclusion $U \hookrightarrow V$ through maps of pairs $(U, U \setminus \{N\}) \to (V, V \setminus \{N\})$. This shows that the local degree of $f_k$ at $N$ is the same as the local degree of the identity at $N$ (since the identity restricts to the inclusion $U \hookrightarrow V$), which is $1$.

This should cover all the technical details; the remainder of the proof follows as described in Hatcher and the other answers.