Just begin to study the CW complexes and not sure if I understand the way the projective plane is constructed as one. In Topology and Geometry by Glen Bredon it is defined as a quotient of $S^n$, which is constructed via attaching 2 "antipodal" cells, hemispheres, in each dimension, under the quotient identifying those hemispheres, which is not the construction of CW complex, but rather a quotient of one.? On the other hand, Allen Hatcher on the page following the definition of CW complex does a similar thing but then spits out its cell complex structure as $e^0\cup e^1\cup e^2\cup\cdots\cup e^n$ and here is where I'm stuck.

In attempt to construct $\mathbb{R}P^1$, following the Hatcher's notations, how is for example the attaching map $\varphi:S^0\longrightarrow X^0$ left total for $X^0=e^0$? That is to get $X^1=e^0\cup e^1$ we should be left with $e^1$ open, am I right?

Out of curiosity, is it at all possible to construct the real projective plane "quotientless", homeomorphic to a pure CW complex?


1 Answer 1


For each $n\geq 0$ define $\mathbb{R}P^n$ as the quotient of $S^n$ by the antipodal $\mathbb{Z}_2$-action, and denote its points $[x_0,\dots,x_n]$, where $(x_0,\dots,x_n)\in S^n\subseteq\mathbb{R}^{n+1}$. For $m<n$ we can identify $\mathbb{R}P^m\subseteq\mathbb{R}P^n$ as the subspace consisting of the points $[x_0,\dots,x_m,0,\dots,0]$, and this gives a filtration $\ast=\mathbb{R}P^0\subseteq\dots\subseteq\mathbb{R}P^m\subseteq\mathbb{R}P^{m+1}\subseteq\dots\mathbb{R}P^n$.

Now the quotient maps $\gamma_m:S^m\rightarrow \mathbb{R}P^m$ are compatible with this filtration, in that the inclusion $S^m\subseteq S^n$ covers the inclusion $\mathbb{R}P^m\subseteq\mathbb{R}P^n$, and are obvious candidates for attaching maps. Define


by writing points of $D^m$ as $(\underline x,t)$, where $\underline x=(x_0,\dots,x_m)\in S^m$ and $t\in[0,1]$, with $(\underline x,0)\sim (\underline x',0)$ for all $\underline x,\underline x'\in S^m$, and setting

$$\Gamma_{m+1}(x_0,\dots,x_m,t)=[t\cdot\underline x,\sqrt{1-t^2}].$$

Observe that the restriction to the boundary satisfies $\Gamma_{m+1}|_{S^m}=\gamma_m$, and that the restriction $\Gamma_{D^{m+1}-S^m}$ maps the interior 1-1 onto $\mathbb{R}P^{m+1}-\mathbb{R}P^m$ (we have a canonical representative with $x_{m+1}>0$ for each coset in its image).

We conclude that $\mathbb{R}P^{m+1}$ is obtained from $\mathbb{R}P^m$ by attaching an $(m+1)$-cell along the quotient $\gamma_m:S^m\rightarrow\mathbb{R}P^m$, and since, as noted above, all the maps are compatible with the natural inclusions, we can conclude that $\mathbb{R}P^n$ is a CW-complex with a single cell in each dimension $0\leq m\leq n$, and with $m$-skeleton \mathbb{R}P^m$.

In remains is to address your queries about the first stages. We have that $\mathbb{R}P^0=S^0/\mathbb{Z}_2=[1,0,\dots,0]$ is a single point, and we have defined the characteristic map of the 1-cell by $\Gamma_1(x_0,t)=[tx_0,\sqrt{1-t^2},0,\dots,0]$. The open 1-cell $e^1$ then consists of the points $\{[x_0,x_1,0,\dots,0]\in\mathbb{R}P^n\mid x_1\neq 0\}$. The preimage of this by the quotient map $\gamma_n$ is the open 1-disc $\mathring D^1\subset S^2\subset S^n$ that forms the upper hemisphere of $S^2$. This is an open subset in $S^n$, and since $\mathbb{R}P^n$ carries the quotient topology by definition we conclude that it is open in $\mathbb{R}P^n$

  • $\begingroup$ This is a beautiful construction in a most straightforward notation! Not just for real case, but it also helps to understand a great deal about why $\mathbb{C}P^n$ are built the way they are. Thank You! $\endgroup$
    – apo
    Apr 1, 2018 at 10:42

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