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I'm trying to show there is no injection $f$ between the 2-torus $ S^1 \times S^1$ ( Henceforth " The Torus") and the 2-sphere $S^2$

My ideas so far:

  1. The Torus is compact, the 2-sphere is Hausdorff. Then $f: T^2 \rightarrow f(T^2) \subset S^2 $ is a homeomorphism, implying the Torus can be embedded in the 2-Sphere. not clear that's enough.(It's been a while since I saw this material. Hope I'm not too far off)

This question :No continuous injective function from 2-sphere to torus is similar but goes in the opposite direction

  1. As a map $f: T^2 \rightarrow S^2, f$ is an element of $\pi_2(T^2)$, which by Kunneth is trivial: ( Edit: As pointed out, this is not the case; rather the other way round; it's the group of homotopy classes of maps from S^n into X )

$\pi_2 (T^2)=\pi_2(S^1 \times S^1)= \pi_2(S^1)\times \pi_2(S^1)=0$ (*)

So that every map in $\pi_2(T^2)$ is homotopically trivial. But I can't tell if any such injective map violates this triviality. Thanks for your suggestions.

Can this be done without Invariance of Domain? I vaguely remember it, but I never got a good feel for it.

*Up to some type of isomorphism.

MSIS
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    Do you mean, no continuous injection? – FShrike Aug 31 '22 at 18:05
  • @FShrike: Yes, thanks, let me edit. – MSIS Aug 31 '22 at 18:06
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    A map $f: T^2 \to S^2$ does not correspond to an element in $\pi_2(T^2)$: the elements of that group come from maps $S^2 \to T^2$. Also, the Kunneth theorem typically refers to the behavior of homology with respect to products, not to the behavior of homotopy groups with respect to products. – John Palmieri Aug 31 '22 at 19:57
  • Ah, anonymous downvotes again. Good to be back with the punishing crowd that expects others to read their minds. – MSIS Sep 01 '22 at 13:46
  • @JohnPalmieri : Corrected the part about the Fundamental group. Still, since $\pi_1 (T^2) $ is Abelian , it's isomorphic to $\pi_1 (T^2)$, so in this sense, Horowicz still applies. – MSIS Sep 04 '22 at 21:16
  • I don't understand the second sentence in your most recent comment. It's certainly true that $\pi_1(T^2) \cong H_1(T^2)$, and it's also true that $\pi_2(T^2) \cong 0$. I don't understand why you are bringing in the Hurewicz theorem or what you mean by "still" applies. "Still" implies that you had already mentioned the Hurewicz theorem, but maybe that's not what you meant. – John Palmieri Sep 04 '22 at 21:44

4 Answers4

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If you could embed the torus in the sphere, then --- since the sphere and the torus are not homeomorphic --- you could embed the torus in the plane. Now consider a parallel and a meridian in the torus, their images in the plane, and remember Jordan's theorem in curves.

Mariano Suárez-Álvarez
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  • You could also directly use the Jordan separation theorem for the $2$-sphere. – Cheerful Parsnip Aug 31 '22 at 19:54
  • I don’t see the problem. You’ve got two circles which intersect at a certain point. They’re being continuously injected into the plane, and Jordan’s theorem gives that the plane is divided into two connected components by both circles. What next? – FShrike Aug 31 '22 at 21:55
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    Invariance of domain implies that the image of the torus in the plane contains an open neighborhoof of the image of the meridian, and thus intersects the two connected components into which the image of the meridian cuts the plane. On the other hand, the image of the complement of the meridian is connected, so it is entirely contained into one of those connected components. – Mariano Suárez-Álvarez Aug 31 '22 at 23:36
  • and without invariance of domain? – psl2Z Sep 01 '22 at 10:08
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Since $S^1\times S^1 \ncong S^2$ this embedding is not surjective and therefore $S^1\times S^1$ embeds in the plane $\mathbb{R}^2$. So we have a subspace of the plane with fundamental group $\mathbb{Z}^2$, which is finitely generated but not free. This is a contradiction. See https://math.stackexchange.com/a/2643521/1016538.

psl2Z
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  • Hi, sorry, dont follow your argument: how does embedding not being surjective imply $S^1 \times S^1$ embeds in the plane $\mathbb R^2$? – MSIS Aug 31 '22 at 19:54
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    @MSIS $S^2\setminus {p} \cong \mathbb{R}^2$ – psl2Z Aug 31 '22 at 19:57
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It's enough to prove that $T^2$ does not embed in $\mathbb R^2.$ But $T^2$ isn't even immersed in $\mathbb R^2.$ For, if so, then for an immersion $f,$ one could find a boundary point $v$ in the image of $f$, (by the compactness of $f(T^2)$). Now, take a $u\in f^{-1}(v)$ and consider $f_*:T_u(T^2)\to T_v(\mathbb R^2).$ The matrix of $f_*$ is the Jacobian $\mathscr J$ of $f$ at the point $u$ (relative to some charts $(U,\varphi), (V,\psi)$ about $u,v,$ respectively). If $\mathscr J$ were invertible, then the inverse function theorem would show that $f$ is a diffeomorphism between some neighborhoods $u\in U'$ and $v\in V'$ in $T^2$ and $\mathbb R^2$, respectively. But this is not possible, because $v$ is a boundary point of $f(T^2).$

Matematleta
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  • I don't think the OP is talking about smooth maps, although I guess you could try to use some of the approximation theorems, but care must be exercised about the properties that are preserved under homotopy. – Laz Aug 31 '22 at 22:49
  • @Laz Yes, I didn't read the question carefully enough. Certainly by Whitney. the OP's $f$ is homotopic to a smooth map. But I think it's even true that that any continuous $f:T^2 \to \mathbb R^n$ can be uniformly approximated by a smooth function. I need to think about this. I will leave my answer up for the present: maybe someone can patch it up. – Matematleta Aug 31 '22 at 23:41
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As already mentioned in other answers it is enough to show that the torus does not embedd in the plane.

Supposing otherwise, assume there is such an embedding $$f:\mathbb T^2 \to \mathbb R^2. $$

Choose a point $p$ in the torus such that $f(p)$ lies in the boundary of $f(\mathbb T^2)$, and let $D$ be a disk in $\mathbb T^2$ centered at $p$. Clearly $f$ restricts to an embedding of $D$ in the plane.

The restriction of $f$ to the boundary of $D$ is therefore a Jordan curve, whose complement in $\mathbb R^2$ is the union of a bounded, connected open set $U$ and an unbounded one, as in the Jordan curve theorem.

Since $f(p)$ is not in the Jordan curve, it must be in $U$ but this contradicts the fact that $p$ lies in the boundary of the range of $f$.

Ruy
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