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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:

  1. 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.
  2. 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.

Shirly Geffen
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    What tools do you have? –  Nov 08 '15 at 14:46
  • Basic Algebraic Topology – Shirly Geffen Nov 08 '15 at 14:48
  • I know one proof , but that requires some strong algebraic topology tools, namely Alexander Duality. – Anubhav Mukherjee Nov 08 '15 at 14:52
  • It is not so clear what's basic algebraic topology mean. Can you briefly describe that? –  Nov 08 '15 at 14:53
  • First degree... – Shirly Geffen Nov 08 '15 at 15:07
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    If you know degree that's enough: you have an embedding of $T^2$ into $S^2$; because it misses a point the local degree formula implies your map is both degree 1 and 0. –  Nov 08 '15 at 15:19
  • @MikeMiller how can you say that this map is degree 1 map? – Anubhav Mukherjee Nov 08 '15 at 15:36
  • @Anubhav.K: An embedding between spaces of the same dimension is a local homeomorpism. Use the local degree formula to see that if $x$ is in the image, the degree is the number of points in $f^{-1}(x)$: one. –  Nov 08 '15 at 15:37
  • @MikeMiller the way Hatcher has defined degree, it says that $f: S^n \to S^n$ – Anubhav Mukherjee Nov 08 '15 at 15:38
  • You can define it for any closed oriented manifold, whose top homology is automatically $\Bbb Z$. The proof of the local degree theorem goes through just fine in this context. –  Nov 08 '15 at 15:39
  • @MikeMiller I need to think about this , I've never thought of such a generalization...in that case your solution looks fine to me, I'd only that much doubt regarding the dedinition of degree – Anubhav Mukherjee Nov 08 '15 at 15:41

3 Answers3

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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 $\mathbb R^2$.

  • @user99914 : Where does that argument fail if you try to embed $T^2$ in $\mathbb R^3 $? – MSIS Sep 13 '22 at 18:43
  • Maybe because IOD requires that $V$ and $F(V)$ live in $\mathbb R^n$ for the same n, I guess. – MSIS Sep 13 '22 at 19:16
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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.

Mariano Suárez-Álvarez
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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.

Lee Mosher
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  • I see, so this does not contradict that the torus can be embedded in $\mathbb R^3$, since, Invariance of domain does not hold between Euclidean spaces of different dimension. – MSIS Sep 16 '22 at 20:13