Is the image of an open ball in $I\times I$ open in the torus via the quotient map? Let $T^2$ be the torus defined as a quotient space $[0,1]^2/\sim$. Consider $a=(1/2,1/2)\in [0,1]^2$ and $r=1/4$.

Prove that image of the open ball $B(a,r)$ via the quotient map $\pi: [0,1]^2\to T^2$ is also an open in $T^2$.

I can imagine the open ball $B(a,r)$ in $[0,1]^2$ as an open subset in $T^2$. However, I can prove it. The quotient map $\pi: [0,1]^2\to T^2$ is not open!
Any idea?
 A: By definition a set is open in the quotient topology if and only if its preimage under the quotient map $x \mapsto [x]$ is open in the original space.
In your case, $T^2$ is the quotient of $[0,1]^2$ by the equivalence relation $(0,a) \sim (1,a), (b,0) \sim (1,b)$ for any $a,b \in [0,1]^2$ (or some equivalent way of phrasing the equivalence relation). So what you should do is:

*

*Find the set of all points equivalent to some point in $B(a,r)$ under this equivalence relation. This will be the preimage of the image of $B(a,r)$ under the quotient map.

*Show that this set is open in $[0,1]^2$.

A: Let $p : I^2 \to T^2$ denote the quotient map. Since $I^2$ is compact and $T^2$ is Hausdorff, it is a closed map.
For each $A \subset I^2$ we have $p(A) \cup p(I^2 \setminus A) = p(A  \cup (I^2 \setminus A)) = p(I^2) = T^2$.
For each $A \subset (0,1)^2$ we have $p(A) \cap p(I^2 \setminus A) = \emptyset$. To see this, assume that $p(x)= p(y)$ with $x \in A$ and $y \in I^2 \setminus A$. This means $x \sim y$. Since nontrivial identifications with respect to $\sim$ are only made between points of the boundary  $B = I^2 \setminus (0,1)^2$ of $I^2$, we must have $x,y \in B$ because $x \ne y$. But $x \notin B$ which gives a contradiction.
Thus for each $A \subset (0,1)^2$ we get
$$p(A) = T^2 \setminus p(I^2 \setminus A) \tag{1}.$$
For an open $A \subset (0,1)^2$ (e.g. $A = B(a,r))$ the complement $I^2 \setminus A$ is closed in $I^2$, thus $p(I^2 \setminus A)$ is closed in $T^2$. Therefore $(1)$ shows that $p(A)$ is open.
