TITLE: Torus cannot be embedded in $\mathbb R^2$
QUESTION [5 upvotes]: I've shown that $T^2$ can be embedded in $\mathbb R^3$.
I just can't see why it can not be embedded in $\mathbb R^2$.
Ideas:

suppose $F: \mathbb S^1\times \mathbb S^1 \to \mathbb R^2$ is continuous injective then we can construct (somehow) $G:\mathbb S^1\to \mathbb R$ continuous injective so we get a contradiction.
we know that $\mathbb R^2$ is homeomorphic to the punctured $2$-sphere, so we get embedding $F$ from $T^2$ to $\mathbb S^2\setminus\{0\}$, thus $F$ is not onto $\mathbb S^2$ and then I can show that it is homotopic to the constant map, but I can't see any contradiction in this situation.

Thank you in advance.

REPLY [10 votes]: Suppose there is an embedding of $T^2$ in the plane. Since the complete graph $K_5$ can be drawn on the torus without crossings, then it follows that it can also be drawn in the plane. But it can't.