I'm stuck with this exercises of my notes:

Let's consider a square $Q=[0,1]\times[0,1]\subset \mathbb{R^2}$, with the usual topology. Let's $p: Q \to T$ be the canonical projection on the torus $T$. Prove that $p$ is not open.

I think that maybe I'm not correct about my way of thinking on the "canonical projection", because if I think that the torus is constructed identifying the sides of that square and doing continuously deformations to it (see the image below), every open set on $[0,1]\times[0,1]$ leads me to an open set on $T$.

Torus from a square

What am I doing wrong?

Thanks for your time.

  • $\begingroup$ It might be easier to view it this way: The torus and square are just two different topologies on the set $[0,1]^2$, and $p$ is the identity map. $\endgroup$ – Henricus V. Jan 21 '18 at 12:52
  • $\begingroup$ @HenricusV. Not exactly. The torus is a topology on $[0,1)^2$ in this interpretation. $\endgroup$ – Arthur Jan 21 '18 at 13:13
  • $\begingroup$ @Arthur If you glue the edge so that the topology is not Hausdorff, this viewpoint is still valid. $\endgroup$ – Henricus V. Jan 21 '18 at 13:14
  • $\begingroup$ @HenricusV. If you mean analogously to the number line with two origins, only one dimension up, sure. It would make $p$ a bijection, but I wouldn't call the result a torus. $\endgroup$ – Arthur Jan 21 '18 at 13:17

Hint: take a small open neighborhood around $(0,0)$.

  • $\begingroup$ A small neighborhood like $[0,\epsilon) \times [0,\epsilon)$? How I prove that the projection image is not open? $\endgroup$ – Relure Jan 21 '18 at 12:04
  • $\begingroup$ I'm sorry Arthur but I think that my problem is understanding how $p$ send the points of the square to the torus $\endgroup$ – Relure Jan 21 '18 at 12:05
  • $\begingroup$ @Relure $p$ can be viewed as the identity map. $\endgroup$ – Henricus V. Jan 21 '18 at 12:51

You're doing nothing wrong, the map does identify the sides like in your picture, except a small neighbourhood of a boundary point in the square gets mapped to only one "half" of an open set.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.