Denote $X=mP^2$ the sphere glued with $m$ Mobius bands. It has a polygon representation $a_1a_1...a_ma_m$, i.e. it's a quotient by a $2m$-sides polygon $P$. Let $o$ be the center point of $P$, $x_0$ a vertex of $P$.

Denote $X_1=P\backslash\{o\}$, then $X_1$ is homotopy to $\partial P$, a bouquet of $m$-circles. So $$ \pi_1(X_1,x_0)=<a_1,...,a_m>, $$

the free group generated by loops $a_1,...,a_m$.

Consider the inclusion map $$ f:\pi_1(P\backslash\{o\},x_0)\to \pi_1(P,x_0), $$ it's a surjective homomorphism. $a_1^2a_2^2...a_m^2\in \pi_1(P\backslash\{o\},x_0)$ is mapped to the trivial elment of $\pi_1(P,x_0)$.

Denote $[a_1^2...a_m^2]$ the normal subgroup generated by $a_1^2...a_m^2$, we know that $[a_1^2...a_m^2]\subset Ker f$. How to prove that $$ Ker f=[a_1^2...a_m^2] $$ without using the conclusion of Van-Kampen theorem? In this simple case, I prefer a direct proof.

isthe space whose fundamental group you are trying to compute. $\endgroup$ – HJRW Mar 15 at 17:52