Let $\mathbb{D}^3=\{(x,y,z)\in \mathbb{R}^3: x^2+y^2+z^2\leq 1\}$. Define $X=\mathbb{D}^3/\sim$ with the relation $(x,y,z)\sim(-x,-y,-z)$ Let $\mathbb{D}^3=\{(x,y,z)\in \mathbb{R}^3: x^2+y^2+z^2\leq 1\}$. Define $X=\mathbb{D}^3/\sim$ with the relation $(x,y,z)\sim(-x,-y,-z)$ for $x,y,z$ such that $x^2+y^2+z^2=1$. Calculate the fundamental group of $X$.
I know that one way to do this is by using Van Kampen's Theorem where we decompose $X=U\cup V$ with $U,V, U\cap V$ connected by paths and in this case I think we could take $U=X-\{\overline{(0,0,0)}\}, V=\pi(\text{Int}(\mathbb{D}^3))$ where $\pi :\mathbb{D}^3\to \mathbb{D}^3/\sim$ is the quotient function, but we have $\pi_1(U,\overline{(1/2,0,0)})=\pi_1(V,\overline{(1/2,0,0)})={1}$ because $\mathbb{S}^2$ is a retraction of deformation of $U,V$, whereby the fundamental group of this space would not be trivial?
 A: $\mathbb{S}^2$ is a deformation retract of $U\cap V,$ not $U$ nor $V.$ Hence $\pi_1(U\cap V,\overline{(1/2,0,0)})=\{1\}.$
$V$ is a ball so it is contractible, hence $\pi_1(V,\overline{(1/2,0,0)})=\{1\}.$ 
Now show that the space
$$ A:=\{(x,y,z)\in \mathbb{D}^3 \text{ | } x^2+y^2+z^2=1\}/\sim \thinspace\subset U$$
is a deformation retract of $U.$ Since $A=\mathbb{R}P^2$ then $\pi_1(U,\overline{(1/2,0,0)})\cong \mathbb{Z}_2.$ Therefore, by Seifert-van Kampen,
$$ \pi_1(X,\overline{(1/2,0,0)})\cong \mathbb{Z}_2.$$
PD: ¿Topología con José Manuel?
A: This is homeomorphic to $\mathbb{R}P^3$, which has $\mathbb{Z}/2\mathbb{Z}$ as fundamental group. 
One way to see why they are homeomorphic is to realize that your quotient is homeomorphic to the quotient of the top closed hemisphere of $S^3$ quotiented in its boundary by $x\leftrightarrow -x$, and this in turn is homeomorphic to the quotient of the whole sphere via the relation $x \leftrightarrow -x$. Each of those homeomorphisms are quite direct to define, and that each one of them is a homeomorphism is straightforward to prove via the universal property of quotients, say.
That it has $\mathbb{Z}/2\mathbb{Z}$ as fundamental group can be seen by noticing that $S^3$ provides a double cover for $\mathbb{R}P^3$. Since $S^3$ is simply-connected, this implies that $\pi_1(\mathbb{R}P^3)$ is a group of order 2. There is only one such group.
