10

I have been trying to prove that there is no embedding from a torus to $S^2$ but to no avail.

I am completely stuck on where to start. The proof is supposed to be based on Homology theory. I know how to prove that $S^n$ cannot be embedded in $\mathbb{R}^n$ however that hasn't helped me in this case. Any help/other eamples of how to prove a lack of an embedding would be great.

Matthew
  • 1,384
  • What do you use to prove $S^n$ cannot be embedded in $\mathbb{R}^n$? Do you know Alexander Duality? – William Apr 03 '19 at 19:09
  • Do you know Alexander duality ? – Maxime Ramzi Apr 03 '19 at 19:11
  • @william I haven't studied Alexander duality so the proof should be able to be done without it. To prove $S^n$ can't be embedded in $R^n$ I assumed it could and then showed that the restricted embedding from $S^n$ to itself not being surjective leads to a contradiction using MV sequence. – Matthew Apr 03 '19 at 19:15
  • Do you know the Invariance of Domain theorem? – Lee Mosher Apr 03 '19 at 22:05
  • @LeeMosher yes I do, however I couldn't figure a way to apply it to the torus – Matthew Apr 03 '19 at 22:08

2 Answers2

6

Suppose that there exists an embedding $f : T^2 \mapsto S^2$.

Each point $x \in T^2$ has an open neighborhood $U$ homeomorphic to the open unit disc in $S^2$, and it follows that $f(U) \subset S^2$ is homeomorphic to the open unit disc. By the Invariance of Domain theorem, $f(U)$ is an open subset of $S^2$. This shows that $f(T^2)$ is an open subset of $S^2$.

But $T^2$ is compact, so $f(T^2)$ is compact, so it is also a closed subset of $S^2$.

By connectivity of $S^2$, it follows that $f(T^2)=S^2$. So $f$ is a homeomorphism, contradicting that $S^2$ is simply connected and $T^2$ is not.

Lee Mosher
  • 135,265
  • Good answer. I have a question, I'm confused how you went from saying that for each point of the torus that $f(U)$ is an open sunset to saying that the image of the torus is. Are you taking the union of all those U for every point of the torus? Thanks – Matthew Apr 04 '19 at 07:09
  • 1
    I realized this argument is much more general than just $T^2$ and $S^2$, it shows that if $M$ and $N$ are manifolds of the same dimension where $M$ is closed and connected, then an embedding $M\to N$ is a homeomorphism onto a component of $N$. – William Apr 04 '19 at 12:14
  • 1
    The reason $f(T^2)$ is an open subset of $S^2$ is that it is a union of open subsets of $S^2$: for every point $y \in f(T^2)$ there exists an open subset $V \subset S^2$ such that $y \in V \subset f(T^2)$. To specify $V$, choose $x \in T^2$ such that $f(x)=y$, use the $U$ given in my answer, and let $V=f(U)$. @Matthew – Lee Mosher Apr 04 '19 at 14:56
  • @William: Yup ! – Lee Mosher Apr 04 '19 at 14:57
4

Here are given several reasons why the torus cannot be embedded into $\mathbb R^2$; two of them use the invariance of domain theorem.

Now, if the torus could be embedded into $S^2$, then this embedding cannot be onto $S^2$, as otherwise this would be a homeomorphism. Thus, as $S^2$ minus a point is homeomorphic to $\mathbb R^2$, we would get an embedding of the torus into $\mathbb R^2$.

Camilo Arosemena Serrato
  • 9,967