Find the fundamental group of $X$ Let $X$ be the unit square with corners identified. I was thinking about its fundamental group. My strategy was to visualize it as e CW complex with a single $0$-cell, four $1$-cells (i.e a wedge of four circles) and a $2$-cell attached along the $1$-cells. This will create a relation and $\pi_1(X) = \mathbb{Z} *\mathbb{Z} *\mathbb{Z}$.
Is this correct?
 A: [Note: I don't have enough reputation points (yet) to include images]
You have the unit square, with edges identified as $A, B, C$, and $D$, respectively, along with vertices $v, v, v,$ and $v$. We will call the square under this identification $X$. Let $U$ be an interior disk of the unit square, let $V$ be the 'boundary' of the unit square such that $U \cap V$ is an annulus intersection in the interior of the square. $U$ is contractible and $V$ deformation retracts onto its edges, which is homeomorphic to the wedge of four circles. Then $U \cap V$ has fundamental group isomorphic to $\mathbb{Z}$, as it has $S^1$ a deformation retract. We apply Siefert-Van Kampen directly, yielding $\pi_{1}(X) \cong \{\alpha, \beta, \gamma, \delta \space | \space \alpha\beta\gamma\delta = 1\} \cong \mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ with the given quotient.
A: Here's another way to see that $\pi_1(X)$ is isomorphic to the free group with three generators. First of all, by thickening the unique $0$-cell in $X$, one see that $X$ is homotopic equivalent to the sphere minus four points. Then using the stereographic projection, this is homeomorphic to $\mathbb R^2$ minus three points. Thus it's fundamental group is $\mathbb Z*\mathbb Z * \mathbb Z$.
