Consider the following quotient space: Take $a,b \in RP^2$, $Y = RP^2 \times \{1, 2, 3\}$, and $X$ = the quotient space of $Y$ quotient by $(b,1) = (a,2)$, $(b,2) = (a,3)$, $(b,3) = (a,1)$. How do we calculate the fundamental group of $X$?
I can calculate it by first calculating the fundamental group of the wedge sum of two real projective planes using Van-Kampen. Then use Van-Kampen again for the three of them together. Is there easier way to do this?
Best Answer
As a matter of clarity, I denote $(a,i)$ and $(b,i)$ by $a_i$ and $b_i$ for $i=1,2,3$, respectively.
Lets call $Z$ the quotient space obtained from $Y$ by identifying $b_1$ with $a_2$ and $b_2$ with $a_3$. First we can compute the fundamental group of $Z$. By applying Van Kampen twice, we get $$\pi_1(Z)\simeq\langle c_1,c_2,c_3\vert c_1^2=c_2^2=c_3^2=1\rangle,$$ where $c_i$ corresponds to the generator of the fundamental group of $RP^2\times \{i\}$.
Now $X$ is obtained from $Z$ by identifying $a_1$ and $b_3$. Let $$p:Z\to X$$ be the corresponding quotient map and let $\overline{a_1}=p(a_1)=p(b_3)$. The map $p$ induces a morphism $$p_*:\pi_1(Z,a_1)\longrightarrow \pi_1(X,\overline{a_1}).$$ We would be happy to say that $p_*$ is onto, but it's not. This is because if $\tilde{\gamma}:[0,1]\to Z$ is a path from $a_1$ to $b_3$, its image $\gamma=p(\tilde{\gamma})$ is a loop in $X$, but it won't be in the image of $p_*$. Let $t$ be some object (I just need $t$ to be some object generating a copy of $\Bbb Z$), there is a natural homomorphism $$f:\langle t\rangle\longrightarrow \pi_1(X,\overline{a_1})$$ sending $t$ to $[\gamma]$. Then $p_*$ and $f$ induce a morphism $$\psi:\pi_1(Z,a_1) * \langle t\rangle\longrightarrow \pi_1(X,\overline{a_1}).$$ I claim that $\psi$ is an isomorphism. If you admit this (which is far from trivial, but reasonable if you picture what $X$ looks like) you get $$\pi_1(X)\simeq\langle c_1,c_2,c_3,t~\vert~ c_1^2=c_2^2=c_3^2=1\rangle.$$
I know that $\psi$ is an isomorphism from a quotient version of Van Kampen that I learned here (unfortunately it is in French). In this reference they also call it the HNN-version of Van Kampen but I haven't found any reference to this specific theorem by searching with these words. If you want to prove that $\psi$ is an isomorphism, you can do something like this:
$\psi$ is onto: Take $U_1=X-\{\overline{a_1}\}$ and $U_2=p(V_1\cap V_2)$ where $V_1$ and $V_2$ are two open sets of $Z$ that deform retract onto $a_1$ and $b_3$, respectively. $U_1$ and $U_2$ are two open sets that cover $X$. Take a path $\alpha:[0,1]\to X$. Then you can decompose $\alpha$ as the concatenation $$\alpha=\alpha_1\cdot\alpha_2\cdots\alpha_{n-1}\cdot\alpha_n$$ such that $\alpha_i$ is a path in $U_1$ or in $U_2$. Then use the fact that $p:Z-\{a_1,b_3\}\to U_1$ is an homeomorphism to express each path $\alpha_i$ with image in $U_1$ as a combination of $\gamma$ and of elements of $\pi_1(Z)$. This kind of technique is very similar to the proof of the classical theorem of Van Kampen.
$\psi$ is one-to-one: You can use the covering of $X$ by $$\tilde{X}=RP^2\times \Bbb Z/_{b_i\sim a_{i+1}}.$$ Take some combination of $\gamma, c_1, c_2$ and $c_3$ which is trivial in $X$. You can picture what the lift will look like in $\tilde{X}$ in terms of the coefficient of the combination, and you should understand that the combination must be trivial because the lift is trivial in $\tilde{X}$.
I hope this helps! I didn't add much detail at the end but I can add some later if you want. I can add drawings to clarify the situation too if you'd like.