# No Embedding From A Torus to A Sphere

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.

• What do you use to prove $S^n$ cannot be embedded in $\mathbb{R}^n$? Do you know Alexander Duality? – William Apr 3 at 19:09
• Do you know Alexander duality ? – Max Apr 3 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 3 at 19:15
• Do you know the Invariance of Domain theorem? – Lee Mosher Apr 3 at 22:05
• @LeeMosher yes I do, however I couldn't figure a way to apply it to the torus – Matthew Apr 3 at 22:08

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.

• 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 4 at 7:09
• 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 4 at 12:14
• 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 4 at 14:56
• @William: Yup ! – Lee Mosher Apr 4 at 14:57

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 $$\mathbb R^2$$.