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Check or falsify: There is a $C^\infty$-map $S^1 \mapsto S^1 \times S^1$ that is surjetive.

Comments: I think it's not true, because $\dim S^1 = 1$ and $\dim S^1 \times S^1 = 2$ but I do not know how to justify.

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  • $\begingroup$ I'm assuming you want the map to be smooth, since you use the differential geometry tag and talk about dimension. Do you know Sard's theorem? If so, then your dimension counting argument applies. $\endgroup$ – Aweygan May 1 '18 at 2:41
  • $\begingroup$ The observation that dim $S^1 \times S^1 > $ dim $S^1$ is not sufficient to rule out the existence of a surjective map: see en.wikipedia.org/wiki/Space-filling_curve. $\endgroup$ – Lucy Yang May 1 '18 at 2:41
  • $\begingroup$ @Aweygan As far as I can tell from the question, the map is not even required to be continuous, let alone differentiable. $\endgroup$ – Lucy Yang May 1 '18 at 2:44
  • $\begingroup$ @Aweygan I know Sard's Theorem, but I do not know how to apply it. $\endgroup$ – Pampas May 1 '18 at 2:56
  • $\begingroup$ @LucyYang for those reasons, I've voted to close the question. I dont think it's clear what category OP is working in, so no answer is likely to be appropriate. Outside of the smooth (or manifold) setting, it's not even clear what $\mathrm{dim}$ means, since I doubt this is topological dimension. $\endgroup$ – Andres Mejia May 1 '18 at 3:05
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I assume you mean to check whether or not there is a smooth map $S^1\to S^1\times S^1$.

In this case, the answer is no. Suppose $f:S^1\to S^1\times S^1$ is surjective. By Sard's theorem, there is a regular value $y\in S^1\times S^1$ of $f$. Since $f$ is surjective, $f(x)=y$ for some $x\in S^1$. But then $df_x$ is surjective, so that $$1=\dim T_xS^1=\dim\ker df_x+\dim T_{y}(S^1\times S^1)\geq\dim T_{y}(S^1\times S^1)=2,$$ a contradiction.

Note that this argument applies in greater generality: If $N^n$ and $M^m$ are smooth manifolds with $m<n$, then any smooth map $f:M\to N$ is not surjective.

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Aweygan showws there is no smooth surjective map, while Danny uses cardinality to show there is a surjection $S^1\rightarrow S^1\times S^1$. Since cardinality ignores things like the topology, the kinds of surjections that Danny is talking about are typically horribly discontinuous.

These leaves open an intermediate question: is there a continuous surjective map from $S^1$ onto $S^1\times S^1$? The answer is that there are continuous surjective maps. Here is one way of building them.

Start with a space-filling curve, that is, a surjective continuous map $[0,1]:I\rightarrow [0,1]\times [0,1]$. The topological space $[0,1]\times [0,1]$ has a natural surjective continous map to $S^1\times S^1$, given by identifying the boundary in the usual way to get a torus. Composing gives a continuous surjection $f:[0,1]\rightarrow S^1\times S^1$. We will imagine the domain of $f$ as the northern hemisphere of $S^1$.

Now, choose any continuous path $\gamma:[0,1]\rightarrow S^1\times S^1$ which connects $f(0)$ to $f(1)$. We will think of the domain of $\gamma$ as the southern hemisphere of $S^1$.

Then, define $F:S^1\rightarrow S^1\times S^1$ by $$F(x) = \begin{cases} f(x) & x \text{ is in northern hemisphere}\\ \gamma(x) & x \text{ is in southern hemisphere}\end{cases}.$$ Because $f$ and $\gamma$ are continuous, and $f(0) = \gamma(0)$ and $f(1) = \gamma(1)$, $F$ is continuous. Because $f$ is surjective, so is $F$.

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  • $\begingroup$ Thanks for posting this. I wanted to post a complete answer considering all three cases, but I didn't have anything to say about the topological case. $\endgroup$ – Aweygan May 1 '18 at 3:36
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No. There is such a map unless we impose restriction on that map. The reason is simple. $S^1$ and $S^1\times S^1$, as sets, have the same cardinality as $\mathbb{R}$. Hence there is a bijection from $S^1$ onto $S^1\times S^1$.

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