Fundamental Group of $\mathbb{RP}^n$

Question

I was tring to culculate the fundamental group of $\mathbb{RP}^n$ with VAN KAMPEN to have a better understanding on how to use this theorem. $\left(\mathbb{RP}^n:=S^n/ \sim \left((x_0,\cdots,x_n) \sim(-x_0,\cdots,-x_n)\right)\right)$

By consider the $A=\left\{[(x_0,\cdots, x_n)] \in \mathbb{RP}^n|-1<x_0<1\right\}$, $B=\left\{[(x_0,\cdots, x_n)] \in \mathbb{RP}^n|x_0>0\right\}$. In case $n=2$ Obviously, we have that the fundamental group of $B$ is $1$ and of $A$ is $\mathbb{Z}$ and consider the element of fundamental groub of $A$ which is $a$. We have that $a^2$ is in the intersection of $A$ and $B$. Thus, $a^2$ shoud be $1$ in fundamental group of $\mathbb{RP}^2$. Therefore the fundamental group of $\mathbb{RP}^2$ shoud be $\mathbb{Z}/2$.

Then I consider the $n>2$ case, I realise that the fundamental group of $B$ is still $1$. And $A$ is the $\mathbb{RP}^{n-1}$. The fundamental group of $A\cap B$ shoud be 1 when $n>2$, because the $A\cap B$ is $S^{n-1}$. I am not sure if I could use the Van Kampen directly to say that the fundamental group of $A\cup B$ is the same as $A$ because the fundamental group of both $A$ and $A\cap B$ is $1$.

Answer 1

Your result for $n=2$ is correct, but the process seems to be unclear to me. In general, we can proceed inductively to find the fundamental group.

The real projective space $\Bbb{RP}^n$ is homeomorphic to $D^n/{(x\sim-x)}$ where $x\in\partial D^n\approx S^{n-1}$. Now, let $U$ be a smaller open $n$-ball $\subset D^n$, then $\pi_1(U)=1$. Let $V$ be an enlargement of $D^n\setminus A$ (open), then $V\simeq S^{n-1}/(x\sim-x)$ by deformation retraction. You may find that $V\approx \Bbb{RP}^{n-1}$ because the most typical definition for real projective spaces is $S^n/(x\sim-x)$.

By Seifert Van-Kampen's Theorem, we conclude that $\pi_1(\Bbb{RP}^n,x_0)\cong \pi_1(U)*\pi_1(V)/N$, but $\pi_1(U\cap V)=1\implies N=e$ (for the reason that $U\cap V\approx S^n, n\ge 2$ is simply connected), so this reduces to $\pi_1(\Bbb{RP}^n)\cong\pi_1(V)$. Because $V\simeq\Bbb{RP}^{n-1}$, $\pi_1(\Bbb{RP}^{n})\cong\pi_1(\Bbb{RP}^{n-1})$. Inductively, it should be $\Bbb{Z}/2$.

However, the case is different when $n=1,2$ because $\Bbb{RP}^1\approx S^1$ and hence has a fundamental group isomorphic to $\Bbb{Z}$ and $\Bbb{RP}^2$ doesn't follow the induction due to the reason that $\pi_1(U\cap V)\cong \Bbb{Z}$.

In sum, $$\pi_1(\Bbb{RP}^n)\cong \begin{cases} \Bbb{Z} & n=1\\ \Bbb{Z}/2 & n\ge2 \end{cases}$$

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Remark: I explored the last statement made by Ted Shifrin in details.

Answer 2

Your writing is way too sloppy, confusing spaces and their fundamental groups. And you really need to consider the image of $\Bbb Z\cong\pi_1(A\cap B)\to\pi_1(A)\cong\Bbb Z$. Van Kampen says you mod out by that image. That gives you $\pi_1(\Bbb RP^2)\cong\Bbb Z/2\Bbb Z$.

In general, you proceed by induction. You get that $\pi_1(\Bbb RP^n) \cong\pi_1(\Bbb RP^{n-1})$ for $n\ge 3$, because $\pi_1(A\cap B) =\{1\}$.

Answer 3

Another proof using induction over Van Kampen would be the following:

1. For $n=1$ we have: $\mathbb{RP}^1 \simeq \mathbb{S}^1$ and therefore $\pi_1(\mathbb{RP}^1)=\pi_1(\mathbb{S}^1)=\mathbb{Z}$.

2. For $n=2$ (induction basis) we have: $\mathbb{RP}^2 \cong \mathbb{RP}^1 \sqcup_{\mathbb{S}^1}\mathbb{D}^2$ and therefore $\pi_1(\mathbb{RP}^2)=\mathbb{Z}\ast_\mathbb{Z}T =\mathbb{Z}/2\mathbb{Z}$ where T is the trivial group. This follows from Van Kampen with $U = [\mathbb{RP}^2]$, $V = [\mathbb{D}^2]$ and $U\cap V = [\mathbb{S}^1]$, all seen as equivalence classes in $\mathbb{RP}^1 \sqcup_{\mathbb{S}^1}\mathbb{D}^2 = \mathbb{RP}^1 \sqcup\mathbb{D}^2/\sim$.

Here the equivalence relation $\sim$ identifies the image of the inclusion of $\mathbb{RP}^1$ in $\mathbb{RP}^1 \sqcup\mathbb{D}^2$ with the image of the inclusion of $\mathbb{D}^2$ in $\mathbb{RP}^1 \sqcup\mathbb{D}^2$. These sets are open since the images in $\mathbb{RP}^1 \sqcup\mathbb{D}^2$ are open by definition. Also, they are path-connected as continuous images of path-connected sets.

3. For $n\geq3$ (induction step) we have: $\mathbb{RP}^n \cong \mathbb{RP}^{n-1} \sqcup_{\mathbb{S}^{n-1}}\mathbb{D}^n$ and therefore $\pi_1(\mathbb{RP}^n)=\mathbb{Z}/2\mathbb{Z}\ast_TT =\mathbb{Z}/2\mathbb{Z}$ by a similiar argument as above.