# Prove that the projection from a square to a torus is not open

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 $T$.

What am I doing wrong?

• 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. – Henricus V. Jan 21 '18 at 12:52
• @HenricusV. Not exactly. The torus is a topology on $[0,1)^2$ in this interpretation. – Arthur Jan 21 '18 at 13:13
• @Arthur If you glue the edge so that the topology is not Hausdorff, this viewpoint is still valid. – Henricus V. Jan 21 '18 at 13:14
• @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. – Arthur Jan 21 '18 at 13:17

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