Atlas of $T^2$ with two charts Let $T^2=S^1\times S^1$ be the two-dimensional torus. This is a manifold. There are many ways to see this. One is to view $T^2$ as the product of two manifolds, i.e., two $S^1$. We wish to find an atlas with as few charts as possible. I construct one with three charts. An exercise in the textbook asks whether we can find an atlas with two charts and what if the sets in the charts are simply connected.
I am not sure how to solve this. I guess the problem gives an implication that the first one is yes, the second, no.
 A: Every closed 2-manifold $M$ that can be covered by two simply connected charts is homeomorphic to $S^2$.
I do know some fancy language for this fact: the Lyusternik-Schnirelman category of every closed, connected 2-manifold except $S^2$ is $\ge 3$.
Despite the fancy language, I do not know of any "slick" algebraic topology proof of this fact (there might nonetheless be one, following the hint of @codeofsilence). But I do know a grubby geometric topology proof, which comes down to the Alexander trick.
Suppose $M = U_1 \cup U_2$ where $U_1,U_2$ are simply connected, hence each is homeomorphic to $\mathbb R^2$. Fix a homeomorphism $h : \mathbb R^2 \to U_2$. Consider the frontier of $U_1$, namely the set
$$F_1 = \overline{U_1} \cap \overline{M-U_1} 
$$
Note that $F_1$ is a compact subset of $U_2$, and so there exists $r>0$ such that $h(B(O,r)) \supset F_1$, where $B(O,r) \subset \mathbb R^2$ is the open ball of radius $r$ around the origin $O \in \mathbb R^2$.
Let $I_t : \overline B(O,r) \to \overline B(O,r)$ be an isotopy of the closed ball (defined for $0 \le t < \infty$) that fixes the origin and each point on the boundary, and that has the property that for $x \in B(O,r)$ we have $I_t(x) \to O$ as $t \to \infty$. In other words, $I_t$ pulls all points in the open ball $B(O,r)$ towards the origin.
Next, define an isotopy $J_t : M \to M$, $t \in [0,\infty)$ as follows:
$$J_t(p)=
\begin{cases}
p &\quad\text{if $p \not\in h(B(O,r))$} \\
h(I_t(h^{-1}(p)) &\quad\text{if $p \in h(B(O,r))$}
\end{cases}
$$
This isotopy has the property that $\bigcup_{t \ge 0} J_t(U_1) = M - h(O)$. We can then choose values $t_1 < t_2 < t_3 < \cdots$ diverging to $\infty$ so that
$$J_{t_1}(U_1) \subset J_{t_2}(U_1) \subset \cdots
$$
This shows that the noncompact manifold $M-h(O)$ is a nested union of open subsets homeomorphic to $\mathbb R^2$. It follows that $M-h(O)$ itself is homeomorphic to $\mathbb R^2$, and so $M$ is homeomorphic to the 1 point compactification of $\mathbb R^2$ which is $S^2$ (this, basically, is the Alexander Trick).
A: Hint
You can do two charts by gluing together two cylinders.  But they won't be simply connected,  because cylinders have the homotopy type of the circle.
In fact,  if you take a simply connected piece out of $T^2$, what's left won't be simply connected.   You can probably make this rigorous with Van Kampen.
Since the intersection won't be connected, in general,  you should work with the fundamental groupoid on a set of base points (as is done for $S^1$).
