Is there a continuous open surjective map from the 2-sphere to the 2-torus? Is there a continuous open surjective map from the 2-sphere $S^2$ to the 2-torus $S^1 \times S^1$?
[Some thoughts: Since both spaces are compact, any continuous surjective map is a quotient map.  There are many such maps, but not all of them are open.  Consider projecting the 2-sphere to a disk, distorting the disk into $[0,1]^2$ and then mapping $[0,1]^2$ to the 2-torus (this last mapping is not open).]
 A: If possible, let $f\colon \Bbb S^2\to \Bbb S^1\times \Bbb S^1$ be a continuous open map. Note that $\Bbb Z^2$ acts on $\Bbb R^2$ properly discontinuously as follows: $$ (m,n)\cdot (x,y):=(x+m, y+n)\text{ for }(x,y)\in\Bbb R^2, (m,n)\in \Bbb Z^2.$$ Let $p\colon \Bbb R^2\to \Bbb R^2/\Bbb Z^2$ be the orbit map. Then $p$  is a covering map due to properly discontinuous action. Also, $\Bbb R^2/\Bbb Z^2$ is homeomorphic to $\Bbb S^1\times \Bbb S^1$. Since $\pi_1(\Bbb S^2)$ is the trivial group, we have a map $\widetilde f\colon \Bbb S^2\to \Bbb R^2$ such that $p\circ \widetilde f=f$.

Here, we show that $\widetilde f$ is an open map. So, let $U$ be an open subset of $\Bbb S^2$. Then $V:=f(U)$ is open in $\Bbb S^1\times \Bbb S^1$ as $f$ is an open map. Cover $V$ by admissible open subsets $\{V_\alpha:\alpha\in \mathcal A\}$ of the covering map $p\colon \Bbb R^2\to \Bbb S^1\times \Bbb S^1$, i.e., we can write $V_\alpha$ as countable disjoint union of open sets $\left\{V_{\alpha n}\subseteq_\text{open}\Bbb R^2:n\in \Bbb N\right\}$ such that $p\big|V_{\alpha n}\to V_\alpha$ is a homeomorphism for each $(\alpha, n)\in \mathcal A\times \Bbb N$.
Define $U_\alpha:=f^{-1}(V_\alpha)\cap U$, i.e., $\{U_\alpha:\alpha\in \mathcal A\}$ is an open cover of $U$. So, $p\widetilde f(U_\alpha)\subseteq_\text{open} V_\alpha$ for each $\alpha\in \mathcal A$ as $f$ is an open map. Thus  $\widetilde f(U_\alpha)$ is the disjoint union of open sets $\left\{\widetilde f(U_\alpha)\cap p^{-1}(V_{\alpha n})\subseteq_\text{open}\Bbb R^2: n\in \Bbb N\right\}$ for each $\alpha\in \mathcal A$. In particular, $\widetilde f(U_\alpha)$ is open in $\Bbb R^2$ for each $\alpha\in \mathcal A$. Therefore $\widetilde f(U)=\bigcup_{\alpha\in \mathcal A} \widetilde f(U_\alpha)$ is an open subset of $\Bbb R^2$. Thus $\widetilde f\colon \Bbb S^2\to \Bbb R^2$ is an open map; in particular, $\widetilde f (\Bbb S^2)$ is an open subset of $\Bbb R^2$. (Thanks to @Dabouliplop
for pointing out an error in the earlier version of this answer)

Also, $\Bbb S^2$ is compact, and $\Bbb R^2$ is Hausdorff, i.e., $\widetilde f(\Bbb S^2)$ is a closed subset of $\Bbb R^2$. The only clopen subset of the connected space $\Bbb R^2$ is either $\varnothing$ or $\Bbb R^2$. Therefore, $\widetilde f(\Bbb S^2)=\Bbb R^2$, which is impossible as a continuous image of a compact space is compact, and $\Bbb R^2$ is non-compact.
A: I just wanted to provide an alternative proof (which makes a bit more sense to me) that the map $\tilde f$ in @Sumanta's answer is open.
Take $U \subset S^2$ open, so that $V = f(U)$ is open.  Since $p$ is a covering map and $V$ is open, for each $x \in V$ there is a connected open set $V_x \subset V$ containing $x$ such that $p^{-1}(V_x)$ can be written as the union of disjoint open sets $V_{x \beta}, \beta \in \mathcal{B}_x$, such that $p \colon V_{x \beta} \to V_x$ is a homeomorphism:
$$p^{-1}(V_x) = \bigcup_{\beta \in \mathcal{B}_x} V_{x \beta}, \quad V =\bigcup_{x \in V} V_x.$$
Since $V_x$ is connected, each $V_{x \beta}$ is connected, and so $\{V_{x \beta}\}$ are the connected components of $p^{-1}(V_x)$.
Letting $U_x = f^{-1}(V_x) \cap U$, we know
$$U = \bigcup_{x \in V} U_x, \quad \tilde f(U_x) \subset p^{-1}(V_x).$$
Let $U_{x \gamma}$, $\gamma \in \Gamma_x$, be the connected components of $U_x$.  Since $U_x \subset S^2$ is open (and so locally connected), the connected components $U_{x \gamma}$ are each open.  Therefore, each $f(U_{x \gamma})$ is open (since $f$ is open).
Since each $\tilde f(U_{\gamma x})$ is connected and contained in $p^{-1}(V_x)$, there is a $\beta_{\gamma}$ such that $\tilde f(U_{\gamma x}) \subset V_{x \beta_\gamma}$.  Therefore, $p \colon \tilde f(U_{\gamma x}) \to p(\tilde f(U_{\gamma x})) = f(U_{\gamma x})$ is a homeomorphism.  Since $f(U_{\gamma x})$ is open, this implies $\tilde f(U_{\gamma x})$ is open.  Hence, $\tilde f(U_x)$ is open, and so is $\tilde f(U)$.
