Define $S^2:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\text{ and }z\ge 0\}$, and $T:=\{(u,v)\in\mathbb{R}^2:u^2+v^2\le 1\}$. Prove we have the homeomorphism \begin{equation}S^2\cong T,\end{equation} where $S^2$ is equipped with the standard topology and $T$ equipped with the subspace topology inherited from $T$.
Idea: By the definition of homeomorphism, we need to find some function $f:S^2\to T$ whereby $f$ is bicontinuous and bijective. Clearly $S^2$ is the unit sphere, and $T$ is the unit disk. So perhaps we can think of projecting from the sphere to the disk; the way in which to do this is not immediate though. Intuitively we can do this geometrically, by simply taking a point on the sphere's surface and projecting it vertically downwards to a unique point on the disk, but it's not obvious to me how to produce an explicit such homeomorphism. (Of course, this ought to be a homemorphism as it is clearly bijective and bicontinuous).
Sure, just project. Consider $S^2_+\ni(x,y,z)\mapsto(x,y)\in T$. This is a restriction and corestriction of projection, and we know projection is continuous, so this map is continuous.
Moreover, this map is surjective. That's clear: given $(x,y)\in T$, $z:=\sqrt{1-x^2-y^2}$ exists as a positive real number and $(x,y,z)\in S^2_+$, so $(x,y)$ is the image of $(x,y,z)$. Injectivity is similarly seen.
I leave it to you to rigorously conclude this map is a homeomorphism. Hint: maps from compact to Hausdorff spaces have a nice property... EDIT: or much more easily, just specify the inverse explicitly (as above). It is not hard to check the inverse is continuous.
Fwiw this is bad notation to call the space $S^2$, actually this space is not homeomorphic to $S^2$ - the unit sphere - so as a hemisphere I would rather call it $S^2_+$ or something like that.