As part of a proof I am reading it states that the part of the sphere given by:
$S$ = $\{(x,y,z) \in \mathbb{R} : x^{2} + y^{2} + z^{2}=1, x \geq 0, y \geq0, z\geq0\} $
is homeomorphic to the closed disc $\textbf{D}^{1}$. It seems intuitively like this would be the case but I am unsure on how to explicitly prove this. My thoughts are that if the fundamental groups are the same they are homeomorphic, and I know the fundamental group of $\textbf{D}^1$ is trivial but I am not sure how I would go about showing the fundamental group of $S$ is trivial.
Consider the map $(x,y,z)\mapsto(x^2-z^2,y,2xz) $ that homeomorphically maps $S$ to $$T:=\{\,(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1,y\ge 0, z\ge 0\,\},$$ then the map $(x,y,z)\mapsto(x, y^2-z^2,2yz) $ that homeomorphically maps $T$ to $$U:= \{\,(x,y,z)\in\mathbb R^3:x^2+y^2+z^2=1,z\ge 0\,\},$$ finally the homeomorphism $U\to \mathbf D^1$, $(x,y,z)\mapsto (x,y)$. This gives an explicit homeomorphism $S\to \mathbf D^1$. (The motivation behind the first two steps is squaring in $\mathbb C$)