# Showing the $2$-sphere with antipodal points identified is homeomorphic to the upper hemisphere with antipodal points identified.

Let $S_+$ be the closed upper hemisphere of the $2$-sphere $S^2$. We can define an equivalence relation $\sim_+$ on $S_+$ as follows:

$x\sim_+ y\:\:\Leftrightarrow\:\:\begin{cases}x=x^\prime,&\text{ or}\\x,\,x^\prime\in\partial S_+\text{ and }x=-x^\prime.\end{cases}$

In other words, we identify antipodal points on the equator of $S_+$. We also have the equivalence relation $\sim$ on $S^2$ in which $x\sim y$ iff $x=y$ or $x=-y$. I want to show $S^2/\sim$ is homeomorphic to $S_+/\sim_+$. Here is what I have done so far:

First, we observe there is an inclusion $i:S_+\hookrightarrow S^2$ of the upper hemisphere $S_+$ into the whole $2$-sphere $S^2$. We also have the canonical maps to the quotient spaces given by $p:S_+\to(S_+)/\sim_+$ and $\pi:S^2\to S^2/\sim$. The map I am going to use is the map $\pi\circ i:S_+\to S^2/\sim$.

Now, we observe that any line through the origin intersects $S^2$ at two points. If we are in the upper hemisphere, then there is one unique intersection, while on the equator there are two points in the (closed) upper hemisphere. In $S^2/\sim$ we are identifying antipodal points, and for each pair of antipodal points, at least one belongs to $S_+$. It is therefore clear that $\pi\circ i$ is surjective. It is also continuous, since it is the composition of continuous functions.

Next, suppose $x\sim_+ y$, then either $x=y$, or $x,\,y$ are antipodal points on the boundary. In either case, $\pi\circ i(x)=\pi\circ i(y)$. Conversely, if $\pi\circ i(x)=\pi\circ i(y)$, then $x,\,y$ are antipodal points, or equal. Since both belong to the (closed) upper hemisphere, we therefore conclude $x\sim_+ y$. So $\pi\circ i$ respects $\sim_+$. By the universal property, there exists a bijection $\varphi:(S_+)/\sim_+\to S^2/\sim$. It therefore remains to show $\varphi$ has a continuous inverse. I suppose I could just do a similar argument, but starting from the map $q:S^2\to S_+$, but I thought it might be neater to show $\pi\circ i$ is an open map, i.e. sends open maps to open maps.

With this in mind, let $U$ be open in $S_+$. Then am I right in saying that $U$ is simply the open ball of a point of $S_+$ that does not lie on the boundary (the equator)? If so, then $i(U)=U$ is also open in $S^2$ and hence $\pi\circ i(U)$ is open in $S^2/\sim$. It is then certainly true that $\varphi$ is a homeomorphism. Is this all alright?

There's an inclusion map $i:S_+\to S^2$. This induces a continuous map $i':{S_+/\sim}\to {S^2/\sim}$. As you point out, this is bijective. But both sides are compact Hausdorff spaces, and a continuous bijection between compact Hausdorff spaces is necessarily a homeomorphism (prove that it's a closed map).
• Ah yes, of course. That would be much nicer. As an aside, am I correct in saying $\pi\circ i$ is an open map? – BillScroggs Apr 11 '18 at 11:41
• @BillScroggs It's an open map all right. It's not true that each open subset of $S_+$ is an open ball though. – Lord Shark the Unknown Apr 11 '18 at 11:48
• @BillScroggs What am I saying? $\pi\circ i$ is not an open map, just as $i$ isn't. But it is a closed map. – Lord Shark the Unknown Apr 11 '18 at 12:52