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I had an exam this morning, one of the questions asked about the truth of the statement

There is a space $X$ such that $S^1$ is homeomorphic to $X\times X$.

I said that this was false and this was my reasoning... was I correct? (Note, an earlier part of the question asked me to prove $\pi_1(Y\times Z)=\pi_1(Y)\times\pi_1(Z)$, I took this as a hint.)

Suppose for the sake of contradiction that such an $X$ exists. Then $\Bbb{Z}\cong G\times G$ where $G\cong\pi_1(X)$. Let $\Phi$ be the isomorphism.

Let $\Phi(1)=(g_1,g_2)\in G\times G$. Then at least one of $g_1,g_2$ is nonzero, since $\Phi(0)=(0,0)$ (the identity on $G\times G$, and $\Phi$ is injective). Without loss of generality $g_1\ne 0$. Then which element $n\in\Bbb{Z}$ has $\Phi(n)=(kg_1,0)$? Since $$kg_1=\underbrace{g_1+\dots +g_1}_k\in G,$$ and $\Phi$ is surjective, such an $n$ should exist. But this is not possible since $\Phi(n)=(kg_1,0)$ implies (since $\Phi$ is a homomorphism) that $kg_1=ng_1$ and $ng_2=0$, and since $k$ was arbitrary and this need not hold if $k\ne 0$.

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  • $\begingroup$ In your context, do homeomorphism and isomorphism mean the same thing? $\endgroup$
    – Vincent
    Apr 27 '16 at 16:18
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    $\begingroup$ @Vincent They do not: the homeomorphism is between the topological spaces $S^1$ and $X\times X$ (a bijective, continuous map with continuous inverse); the isomorphism is between the groups $\pi_1(S^1)$ and $\pi_1(X)\times \pi_1(X)$ (a bijective group homomorphism) $\endgroup$
    – Szmagpie
    Apr 27 '16 at 16:21
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    $\begingroup$ You're close but I don't think this is completely correct. You're supposed to be finding an $n$, so you can just choose $n=k$ then $kg_1=ng_1$, and you could still have $ng_2=0$ without issue. $\endgroup$ Apr 27 '16 at 16:24
  • $\begingroup$ @AlexMathers yes you are right. Now I am trying to resurrect that proof. Maybe it is that $ng_2=0$ would imply $g_2=0$, since in general, $n=k$ may not be the identity. Then I seek an $m$ such that $\Phi(m)=(g_2,g_1)$. This would need in the first "coordinate" $mg_1=g_2=0$, implying $m=0$. But $\Phi(0)=(0,0)\ne (g_2,g_1)$ since $g_1\ne 0$. $\endgroup$
    – Szmagpie
    Apr 27 '16 at 16:55
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    $\begingroup$ You can also show that $X^2$ minus 2 points is connected (for $X$ uncountable and connected, say) and the circle minus 2 points is disconnected. $\endgroup$ Apr 27 '16 at 17:00
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Your idea is OK, using homotopy groups.

Suppose $\mathbb{Z} = G \times G$ for some (Abelian) group $G$, which must be infinite. Then note that $\{0\} \times G$ and $G \times \{0\}$ are infinite subgroups of $G \times G$ that intersect only in the unit element $\{(0,0)\}$. By the isomorphism that supposedly exists, such subgroups should also exist in the integers.

But any subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ and any two of those have infinite intersection (at least $nm\mathbb{Z}$ for $n\mathbb{Z}$ and $m\mathbb{Z}$), so such subgroups cannot exist.

Using your original approach: suppose $h: \mathbb{Z} \rightarrow G \times G$ is an isomorphism, and $h(1) = (g_1,g_2)$, so that $h(n) = (ng_1, ng_2)$. As $(0,g_2)$ should be an image, $h(k) = (0,g_2)$ for some $k$, but it also equals $(kg_1, kg_2)$ so $kg_1 = 0$, and $kg_2 = g_2$, the latter implies $k=1$ (and then $g_1 = 0$ from the first equation), or $g_2 = 0$. In either case the image of $\mathbb{Z}$ can only be a subset of $G \times \{0\}$ or $\{0\} \times G$, both of which are never all of $G \times G$.

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