I am trying to find the fundamental group of a sphere with 3 points identified. It is homotopy equivalent to a sphere wedged sum with two circles, so its group is the free group of two generators.
But I want to find another approach, which is finding its cell structure. Here's what I think. I start with a 0-cell $x$ and two 1-cells attached their boundary to $x$, say $a$ and $b$. And then I attach the $2$ cell to $aa^{-1}bb^{-1}$.
I have no vigorous proof about this, it's just geometric intuition. So is it correct?
There are probably many ways to do it, but I suggest the following:
Start with three $0$-cells $e_0,e_1,e_2$ and attach three $1$-cells $a,b,c$ to obtain a copy of $S^1$. Now attach two $2$-cells to $S^1$ to obtain a copy of $S^2$. The quotient $X = S^2/\{e_0,e_1,e_2\}$ has the structure of a CW-complex with one $0$-cell $e$, three $1$-cells $a',b',c'$ giving a wedge $\Sigma$ of three circles and two $2$-cells attached to $\Sigma$ in the obvious way.
Then $\pi_1(X)$ is the quotient of $\mathbb Z * \mathbb Z * \mathbb Z$ (with generators $a',b',c'$) by the normal subgroup generated by $a'b'c'$.