Homeomorphism between $S^1×S^1$ and $S^2$ Are the two sets $S^1×S^1$ and $S^2$ homeomorphic? 
I am thinking that they are not but I am not getting any argument for that.
 A: Consider the loop $\{1\}\times \Bbb S^1$ in the torus $\Bbb S^1\times \Bbb S^1$. Suppose there is a homeomorpism $f:\Bbb S^1\times \Bbb S^1\to\Bbb S^2$. Then $f(\{1\}\times \Bbb S^1)$ is loop in $\Bbb S^2$. Now, $\Bbb S^2\backslash f(\{1\}\times \Bbb S^1)$ is disconnected. But, $(\Bbb S^1\times \Bbb S^1)\backslash (\{1\}\times \Bbb S^1)$ is connected.
To show, $(\Bbb S^1\times \Bbb S^1)\backslash (\{1\}\times \Bbb S^1)$ is connected, consider two any points, $(e^{i\theta_k},e^{i\psi_k})\in (\Bbb S^1\times \Bbb S^1)\backslash (\{1\}\times \Bbb S^1)$ for $k=1,2$. Here $0<\theta_1,\theta_2<2\pi$ and $0\leq \psi_1,\psi_2\leq 2\pi$. Now, we have a path from $ (e^{i\theta_1},e^{i\psi_1})$ to $(e^{i\theta_1},e^{i\psi_2})$ and a path from $(e^{i\theta_1},e^{i\psi_2})$ to $(e^{i\theta_2},e^{i\psi_2})$. So  juxtapsoing these paths we have a path from $(e^{i\theta_1},e^{i\psi_1})$ to $(e^{i\theta_2},e^{i\psi_2})$. Hence,  $(\Bbb S^1\times \Bbb S^1)\backslash (\{1\}\times \Bbb S^1)$ is path connected.
Next we have to show, $X:=\Bbb S^2\backslash f(\{1\}\times \Bbb S^1)$ is disconnected. To show this, note that, under stereographic projection we have, $X\simeq \Bbb R^2\backslash L$, where $L$ is a line homeomorphic to $\{f(1, e^{i\psi}):0<\psi<2\pi\}$. But, $\Bbb R^2\backslash L$ is disconnected. So we are done. And this completes the proof.
$$\textbf{Note:--  }\frac{\Bbb S^1\times\Bbb S^1}{\big(\Bbb S^1\times\{1\}\big)\cup \big(\{1\}\times\Bbb S^1\big)}\text{ is homeomorphic to }\Bbb S^2.$$
