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 $A=left{[(x_0,cdots, x_n)] in mathbb{RP}^n|-1<x_0<1right}$, $B=left{[(x_0,cdots, x_n)] in mathbb{RP}^n|x_0>0right}$.
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 $Acap B$ shoud be 1 when $n>2$, because the $Acap B$ is $S^{n-1}$.
I am not sure if I could use the Van Kampen directly to say that the fundamental group of $Acup B$ is the same as $A$ because the fundamental group of both $A$ and $Acap B$ is $1$.

algebraic topology

Kevin.S · Answer

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 $D^n/{(xsim-x)}$ where $xinpartial D^napprox S^{n-1}$. Now, let $U$ be a smaller open $n$-ball $subset D^n$, then $pi_1(U)=1$. Let $V$ be an enlargement of $D^nsetminus A$ (open), then $Vsimeq S^{n-1}/(xsim-x)$ by deformation retraction. You may find that $Vapprox Bbb{RP}^{n-1}$ because the most typical definition for real projective spaces is $S^n/(xsim-x)$.
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 $pi_1(Ucap V)=1implies N=e$ (for the reason that $Ucap Vapprox S^n, nge 2$ is simply connected), so this reduces to $pi_1(Bbb{RP}^n)congpi_1(V)$. Because $VsimeqBbb{RP}^{n-1}$, $pi_1(Bbb{RP}^{n})congpi_1(Bbb{RP}^{n-1})$. Inductively, it should be $Bbb{Z}/2$.
However, the case is different when $n=1,2$ because $Bbb{RP}^1approx S^1$ and hence has a fundamental group isomorphic to $Bbb{Z}$ and $Bbb{RP}^2$ doesn't follow the induction due to the reason that $pi_1(Ucap V)cong Bbb{Z}$.
In sum,
$$pi_1(Bbb{RP}^n)cong
begin{cases}
Bbb{Z} & n=1\
Bbb{Z}/2 & nge2
end{cases}$$

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

Ted Shifrin · Answer

Your writing is way too sloppy, confusing spaces and their fundamental groups. And you really need to consider the image of $Bbb Zcongpi_1(Acap B)topi_1(A)congBbb Z$. Van Kampen says you mod out by that image. That gives you $pi_1(Bbb RP^2)congBbb Z/2Bbb Z$.
In general, you proceed by induction. You get that $pi_1(Bbb RP^n) congpi_1(Bbb RP^{n-1})$ for $nge 3$, because $pi_1(Acap B) ={1}$.

Fundamental Group of $mathbb{RP}^n$

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