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Problem : Show that there doesn't exist a space $X$ such that $X\times X$ is homeomorphic to $S^{2}$, the 2-dimensional sphere.

Is there any solution which doesn't use algebraic topology? I solved this problem by using Kunneth formula and there is another solution using homotopy group $\pi_{2}$. However, it seems that there might be an elementary proof that only uses general topology. If such space exists, we can prove that $X$ is connected and compact. Also, we can show that $X$ is simply-connected. Anything else? Thanks in advance.

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    $\begingroup$ I would be surprised to see an answer that doesn't use algebraic topology. $\endgroup$ – user98602 May 17 '17 at 3:34
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    $\begingroup$ I think we also get $\dim(X) =1$ as all spaces involved are compact metric, so we have an addition formula for products and $\dim$. $\endgroup$ – Henno Brandsma May 17 '17 at 4:04
  • $\begingroup$ @HennoBrandsma What is a definition of dimension? Is $X$ a manifold? $\endgroup$ – Seewoo Lee May 17 '17 at 4:15
  • $\begingroup$ Topological dimension. Defined for all spaces. $\endgroup$ – Henno Brandsma May 17 '17 at 4:17
  • $\begingroup$ Also $X$ is a Peano continuum. So it contains topological copies of $[0,1]$ $\endgroup$ – Henno Brandsma May 17 '17 at 4:37
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Here is an algebraic topology free answer, which uses several other big hammers instead.

Assume for a contradiction that $f:X\times X\rightarrow S^2$ is a homeomorhpism.

Consider the involution $\sigma:X\times X\rightarrow X\times X$ which switches the factors: $\sigma(x_1,x_2) = (x_2,x_1).$ Then $\rho = f\circ \sigma \circ f^{-1}$ is an involution of $S^2$. According to Eilenberg (in a proof which doesn't use algebraic topology), $\rho$ is conjugate to a smooth map $\rho '$ in $Homeo(S^2)$ (Thanks to Balarka for noticing and fixing a previous gap here). By averaging a Riemannian metric on $S^2$ we may assume that $\rho'$ is an isometry on $S^2$. In particular, the fixed point set of $\rho'$ must be an embedded submanifold in $S^2$. Note that the fixed point set of $\rho'$ is homeomorphic to the fixed point set of $\rho$.

Now, the fixed point set of $\sigma$ is the diagonal $\Delta X = \{(x, x)\in X\times X: x\in X\}$ so is clearly homeomorphic to $X$. Since $f$ is a homeomorphism, we see the fixed point set of $\rho$, which is equal to $f(X)$, is homeomorphic to $X$. Thus, we have shown that $X$ itself has the structure of a smooth manifold.

From the classification of compact 1-manifolds (which can be done via Riemannian geometry using the exponential map), it follows that $X$ is homeomorphic to $S^1$. Therefore, $S^2$ is homeomorphic to $S^1\times S^1$.

Now, how do we show this is a contradiction without algebraic topology? First, equip $S^2$ and $S^1\times S^1$ with their standard differentiable structures.

In this paper of Hatcher, he shows that any homeomorphism between smooth surfaces is isotopic to a diffeomorphism without using any results from algebraic topology. Thus, we may assume $S^2$ is diffeomorphic to $S^1\times S^1$.

Finally, we note that $S^1\times S^1$ is parallelizable (obvious) while $S^2$ is not (less obvious), so they can't be diffeomorphic. And while the standard proof of the Hairy Ball Theorem uses algebraic topology, Milnor has provided a proof which uses only some analysis and a bit of point set topology.

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    $\begingroup$ This is a really nice proof (+1). I have a small comment to make: $\rho$ need not be a smooth involution, but I think any involution on the 2-sphere is topologically conjugate to a smooth one. On contrast, I think this is false for the 3-sphere (Bing proved that gluing two copies of the exterior of the Alexander horned sphere along the metric boundary by the identity map is homeomorphic to $S^3$; switching the two copies gives an involution of $S^3$ with a non-locally flat submanifold as fixed point locus). Perhaps that kind of thing can cause trouble in your conclusion in the comment above? $\endgroup$ – Balarka Sen Feb 2 '18 at 20:50
  • $\begingroup$ $S^2$ is not homeomorphic to $S^1 \times S^1$ by the Jordan curve theorem, which says a continuously embedded loop must disconnect the sphere. But I agree with Balarka that you are implicitly assuming smoothness here. $\endgroup$ – user98602 Feb 2 '18 at 21:25
  • $\begingroup$ @Balarka: you are right! (Which ruins the corollary in my now-deleted comment) Do you have a reference for the statement that $\rho$ is conjugate to a smooth map? Does the proof avoid algebraic topology? $\endgroup$ – Jason DeVito Feb 2 '18 at 21:51
  • $\begingroup$ @Mike: I agree with the smoothness issue. I thought of using JCT, but for some reason was stuck on the "both pieces are homeo to a disc" part and wondered how I would show this doesn't hold on $S^1\times S^1$ without some algebraic topology. I should have focused on the "there are two pieces" easier part! $\endgroup$ – Jason DeVito Feb 2 '18 at 21:53
  • $\begingroup$ @Balarka: Found a reference. Thanks again! $\endgroup$ – Jason DeVito Feb 2 '18 at 22:24
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Assume it were true. Let $h : X \times X \to S^2$ be a homeomorphism. $X$ must have more than one point, so choose two distinct points $a, b \in X$. $X' = h(X \times \lbrace a \rbrace)$ is a homeomorphic copy of $X$ and a retract of $S^2$ (since $X \times \lbrace a \rbrace$ is a retract of $X \times X$). Since $s = h(a,b) \notin X'$, $X'$ is also a retract of $S^2 \backslash \lbrace s \rbrace$ which is contractible. Therefore $X$ is contractible, hence also $X \times X \approx S^2$. This is a contradiction.

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  • $\begingroup$ I voted this up, but, in retrospect, I'm afraid I don't know how to show that that "$S^2$ is contractible" is a contradiction without invoking some algebraic topology. (The rest of the answer uses terminology from algebraic topology, but I wouldn't say it used any real results from algebraic topology....) $\endgroup$ – Jason DeVito Feb 2 '18 at 19:59
  • $\begingroup$ @JasondeVito The "standard" proof for $S^2$ not being contractible is in fact based on (singular) homology theory. However, there are other proofs (e.g. using Sperner's Lemma which is a purely combinatorial result on simplicial subdivisions of simplices, or using the mapping degree for maps between differentiable manifolds). But all that is non-trivial - in that sense you are right, I can't offer a completely elementary proof. $\endgroup$ – Paul Frost Feb 3 '18 at 10:08
  • $\begingroup$ Concerning Sperner's Lemma I found the following short account: math.mit.edu/~fox/MAT307-lecture03.pdf $\endgroup$ – Paul Frost Feb 3 '18 at 10:12
  • $\begingroup$ Very nice, thank you. I would upvote again if I could ;-) $\endgroup$ – Jason DeVito Feb 5 '18 at 3:32

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