Torus cannot be embedded in $\mathbb R^2$ 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.
 A: If you allow the Invariance of Domain theorem into your algebraic topology toolbox, here's what you can do.
Since $T^2$ is compact, the "radius function" $r(x) = |f(x)|$ has a maximum value $R \ge 0$, achieved at some point $x_M \in T^2$. It follows that $\text{image}(f)$ does not contain any open ball centered on $f(x_M)$. 
Now choose a subset $B \subset T^2$ containing $x_M$ with a homeomorphism between $B$ and the closed unit ball in $\mathbb{R}^2$ such that $x_M$ corresponds to the origin of $\mathbb{R}^2$. The restriction $f | B$ is a continuous bijection onto its image and is therefore a homeomorphism onto its image. 
The restriction $f | \text{interior}(B)$ then violates the invariance of domain theorem: it's domain is homeomorphic to an open ball in $\mathbb{R}^2$, and that restriction is an injective continuous map, but its image contains $f(x_M)$ and contains no point further from the origin than $f(x_M)$ and is therefore not open.
A: Let $F : \mathbb S^1 \times \mathbb S^1 \to \mathbb R^2$. Then the image is closed as $\mathbb S^1 \times\mathbb S^1$ is compact. On the other hand, if $F$ is injective, then the restriction to each small neighborhood $V \subset \mathbb S^1 \times \mathbb S^1$ is also injective. The invariance of domain shows that $F(V)$ is open, and so the image of $F$ is also open. But that is nonsense as the only open and closed nonempty subset in $\mathbb R^2$ is $\mathbb R^2$.   
Note that there is nothing special about the torus. Every compact surface cannot be embedded to $\mathbb R^2$. 
A: 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.
