Showing that these two quotient spaces are not homeomorphic

Question

Let, $P$ be the quotient space obtainted from $S^{2}$ by identifying two distinct points and $Q$ be the quotient space obtained by identifying three mutually distinct points in $S^{2}$. Show that these two spaces are not homeomorphic.

Any suggestions? I'm very confused by the quotient topology section in Munkres and I can't even visualize what these spaces are supposed to "look" like.

Answer 1

Sketch: There is a point $q\in Q$, namely the point to which the three points identify, such that there is a (path-)connected open neighbourhood $U$ of $q$ such that $U\setminus \{q\}$ has three (path-)connected components, whereas for all (path-)connected open sets $V\subseteq P$ and for all $p\in P$, $V\setminus\{p\}$ has at most two (path-)connected components. For the part about $P$, one must mak the case of whether or not $V$ contains the image $c$ of the two points and $p=c$.

If you want a visual embedding in $\Bbb R^3$, grab the two/three points you want to identify and pull the surface of $S^2$ so that those points become the tip of an equal number of little horns. Then stretch said horns and have their tips meet. Depending on the initial number of points, you end up with $P$, $Q$ or possibily something else.

There is of course the inverse construction, where you start with two or three cone-hats having the tip in common and so that no one lies in the interior of the other. Let their bottoms rest on a $S^2$ and join them by attaching the portion of sphere that is outside of all the cones.

Answer 2

Think of the quotient topology like collapsing/gluing subspaces. So in this problem you can visualize P like a balloon which you have pinched from opposite sides until they meet at one point in the middle. It's homeomorphic to the surface in $\mathbb{R}^3$ defined by $$z^2 = (x^2 + y^2)(1-x^2-y^2).$$ Q is the same idea, except you glue three points together.

As for showing that they're not homeomorphic, you just need to find any topological property that is different between them. As in Gae's answer, I would look at small connected neighborhoods of the point of gluing $q$, and then delete the point $q$ from them. For P the resulting neighborhoods will have two connected components, for Q they will have three.