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 $(x,y,z)\sim(-x,-y,-z)$

Question

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?

Answer 1

$\mathbb{S}^2$ 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?

Answer 2

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.