Prove that $\pi$ is a quotient map which is neither open nor closed

continuitygeneral-topologymap-projectionsquotient-spaces

Exercise: Let $X:=\mathbb{R}^2\smallsetminus((-1,1)\times(-2,2)\cup[2,3]\times[-2,2])\subset\mathbb{R}^2$ be equipped with the subspace topology and consider the map $\pi:X\rightarrow\mathbb{R}$, which is the restriction to $X\subset\mathbb{R}^2$ of the projection to the $x$-axis. Show that $\pi$ is a quotient map which is neither open nor closed.

Thoughts: It's straight-forward to see that the map is surjective as it covers all of $\mathbb{R}$ by definition, but I fail to see that it is continuous, and why the map is neither open nor closed. If $X$ were to be equipped with the product topology, then continuity would follow straight from that, but not we are dealing with a projective map with the subspace topology, which stuns me a bit. I also don't really have a clue about not being open nor closed for similar reasons.

Best Answer

  1. Continuity is trivial, as you write, because the restriction of a continuous map to a subspace is always continuous. It does not matter that the map you restrict is a projection, it is still a continuous map. (You can also check this directly, by considering preimages of open intervals on the X axis.)
  2. To see that it is not open, consider the projection of (the intersection with $X$ of) a small open ball around a point on the left side of $(-1,1)\times(-2,2)$.
  3. To see that it's not closed, consider instead a closed ball around a point on the left side of $[2,3]\times[-2,2]$.
  4. To see that it is a quotient map, take the preimage of a subset of the line and notice that if it is open, so is its intersection with e.g. $\mathbf R\times (10,11)\subseteq X$; the restriction of $\pi$ to this set is an open map, which you can use to show that the set you started with had to be open on the line.
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