The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected.

A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this.

So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$?

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    $\begingroup$ I'm wondering to what extent there is unique factorization of topological spaces relative to $\times$. $\mathbb{Q}$ is an idempotent (as is its complement in $\mathbb{R}$), but are there more interesting failures of UF involving connected spaces? Or results establishing UF for "nice" families of spaces? Should these be posted as a new question? $\endgroup$ Apr 3, 2011 at 1:41
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    $\begingroup$ Is Moebius $\times$ Moebius = cilinder $\times$ cilinder (no boundaries)? $\endgroup$ Apr 4, 2011 at 16:38
  • $\begingroup$ Without knowing any algebraic topology, it's possible to conclude at least something about X. If X is metric, compact, or locally compact and paracompact, then $\dim(X\times X)\le 2\dim X$, which means X has to have Lebesgue covering dimension at least 2. Wage, Proc. Natl. Acad. Sci. USA 75 (1978) 4671 , www.pnas.org/content/75/10/4671.full.pdf . What is the weakest condition that guarantees $\dim(X\times Y)= \dim X+\dim Y$? Given Yaakov Baruch's comment about the "dogbone space," it's not obvious that X is at all well behaved simply from the requirement that its square is $\mathbb{R}^3$. $\endgroup$
    – user21349
    Jan 19, 2013 at 15:55
  • $\begingroup$ @YaakovBaruch, isn't the cylinder factorizable? And could you elaborate this identity a little? $\endgroup$
    – Ash GX
    Oct 17, 2013 at 6:35

4 Answers 4


No such space exists. Even better, let's generalize your proof by converting information about path components into homology groups.

For an open inclusion of spaces $X \setminus \{x\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$ H_n(X \times X, X \times X \setminus \{(x,x)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{x\};k) \otimes_k H_q(X, X \setminus \{x\};k). $$ If the product is $\mathbb{R}^3$, then the left-hand side is $k$ in degree 3 and zero otherwise, so something on the right-hand side must be nontrivial. However, if $H_p(X, X \setminus \{x\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.

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    $\begingroup$ I hope this fine illustration of the power of relative homology will find its way in a textbook or, meanwhile, in algebraic topology courses. $\endgroup$ Apr 2, 2011 at 19:40
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    $\begingroup$ I have a question regarding the top answer given by Tyler Lawson. As far as I know you can only apply the relative version of the Kunneth formula to cofibrations. Since we do not know much about $X$, it is unclear why $(X, X\setminus p)$ is a cofibration. Moreover, $(\mathbb R^3, \mathbb R^3\setminus p)$ is not a cofibration (I think). $\endgroup$
    – freddy
    Mar 21, 2017 at 14:21
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    $\begingroup$ For example, Dold's version (Corollary 12.10 in Lectures on Algebraic Topology part VI) requires an excisive triad condition. The core of these assumotions is to ensure that, given $(X,A)$ and $(Y,B)$, the covering of $(X \times B) \cup (Y \times A)$ by $X \times B$ and $A \times Y$ is good enough to satisfy the assumptions of the Mayer-Vietoris theorem. This is, in particular, satisfied if $A$ is an open subset of $X$ and $B$ is an open subset of $Y$, or in the CW-inclusion version that Hatcher uses. $\endgroup$ Mar 21, 2017 at 19:01
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    $\begingroup$ So this also works for $\sqrt{\mathbb{R}^{2n+1}}$ doesn't it? $\endgroup$ Aug 3, 2018 at 8:20
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    $\begingroup$ This should prove that if $\mathbb{R}^n=X^k$, then $k$ divides $n$, I think? $\endgroup$ Oct 17, 2019 at 1:38

this blog post refers to some papers with proofs. I've heard Robert Fokkink explain his proof (which is, quoting from this post)

A linear map $\Bbb R^n \to \Bbb R^n$ can be understood to preserve or reverse orientation, depending on whether its determinant is $+1$ or $-1$. This notion of orientation can be generalized to arbitrary homeomorphisms, giving a "degree" $\deg(m)$ for every homeomorphism which is $+1$ if it is orientation-preserving and $-1$ if it is orientation-reversing. The generalization has all the properties that one would hope for. In particular, it coincides with the corresponding notions for linear maps and differentiable maps, and it is multiplicative: $\deg(f \circ g) = \deg(f)\cdot \deg(g)$ for all homeomorphisms $f$ and $g$. In particular (fact 1), if $h$ is any homeomorphism whatever, then $h \circ h$ is an orientation-preserving map.

Now, suppose that $h : X^2 \to \Bbb R^3$ is a homeomorphism. Then $X^4$ is homeomorphic to $\Bbb R^6$, and we can view quadruples $(a,b,c,d)$ of elements of $X$ as equivalent to sextuples $(p,q,r,s,t,u)$ of elements of $\Bbb R$.

Consider the map $s$ on $X^4$ which takes $(a,b,c,d) \to (d,a,b,c)$. Then $s \circ s$ is the map $(a,b,c,d) \to (c,d,a,b)$. By fact 1 above, $s \circ s$ must be an orientation-preserving map. But translated to the putatively homeomorphic space $\Bbb R^6$, the map $(a,b,c,d) \to (c,d,a,b)$ is just the linear map on $\Bbb R^6$ that takes $(p,q,r,s,t,u) \to (s,t,u,p,q,r)$. This map is orientation-reversing, because its determinant is $-1$. This is a contradiction. So $X^4$ must not be homeomorphic to $\Bbb R^6$, and $X^2$ therefore not homeomorphic to $\Bbb R^3$.

and there he also told us the cohomological proof, which generalizes it to all Euclidean spaces of odd dimension.

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    $\begingroup$ I hope no one misses this nice alternative proof because it's behind a link. $\endgroup$ Apr 4, 2011 at 2:24
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    $\begingroup$ Quoting from the link: "The paper also refers to an earlier paper ("The cartesian product of a certain nonmanifold and a line is E4", R.H. Bing, Annals of Mathematics series 2 vol 70 1959 pp. 399–412) which constructs an extremely pathological space B, called the "dogbone space", not even a manifold, which nevertheless has B × R^3 = R4." This is relevant to my comment to the OP. $\endgroup$ Apr 4, 2011 at 5:16
  • $\begingroup$ I don't understand this step in the proof: Why does the map $X^4 \to X^4, (a,b,c,d) \mapsto (c,d,a,b)$ correspond to the map $R^6 \to R^6, (p,q,r,s,t,u) \mapsto (s,t,u,p,q,r)$? I mean, the homeomorphism is not supposed to commute with projections ... $\endgroup$ Apr 4, 2011 at 15:05
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    $\begingroup$ @Martin: The homeomorphism $(X\times X)\times (X\times X)\cong \mathbb R^3 \times \mathbb R^3$ respects projections by construction, so swapping the "two factors" (which I've emphasized with parentheses) on the left hand side corresponds to swapping the two factors on the right hand side. $\endgroup$ Apr 5, 2011 at 5:42
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    $\begingroup$ This argument is also given as exercise in Hatcher's "More exercises in algebraic topology". $\endgroup$ Oct 6, 2020 at 18:21

I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.

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    $\begingroup$ Would you care to elaborate? $\endgroup$
    – Ian Agol
    Jan 19, 2013 at 5:09
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    $\begingroup$ All things considered, perhaps "S" is not the best name for the topological space for this assertion. $\endgroup$
    – Terry Tao
    Jan 19, 2013 at 5:52
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    $\begingroup$ @Terry Tao True enough, but in all honesty it's precisely the notational perversity that brought this to mind to begin with. $\endgroup$ Jan 19, 2013 at 10:28
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    $\begingroup$ @Agol Fix $s\in S$. On the one hand, $\pi_2(S\times S,(s,s))\cong \pi_2(S,s)\times\pi_2(S,s)$. On the other hand, $\pi_2({\bf S},{\bf s})\cong{\mathbb Z}$ for any 2-sphere $\bf S$ and any ${\bf s}\in{\bf S}$. Now it suffices to observe that ${\mathbb Z}\not\cong G\times G$ for any group $G$: indeed, such a group must be an infinite quotient of $\mathbb Z$, whence $G\cong{\mathbb Z}$, but ${\mathbb Z}\not\cong{\mathbb Z}\times{\mathbb Z}$ $\endgroup$ Jan 19, 2013 at 11:27
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    $\begingroup$ @IanAgol : On the LHS, $S^2$ refers to the $2$-sphere, while on the LHS $S$ refers to an arbitrary topological space. $\endgroup$
    – Prateek
    Nov 17, 2014 at 15:55

The Euler characteristic with compact support $\chi_c(X)$ is a very robust topological invariant available for any reasonable space (such as a subanalytic set). The key properties here are that $\chi_c(X)$ is a real number and $\chi_c(A\times B)$ is multiplicative:

$$ \chi_c(X)\in \mathbb{R},\quad \quad \chi_c(A\times B)=\chi_c(A)\cdot \chi_c(B). $$

It follows that the Euler characteristic with compact support of a topological square is always non-negative: $$\forall X: \quad \chi_c(X\times X)\geq 0.$$

Thus, a space with negative Euler characteristic with compact support cannot be a topological square.

In particular, since $\chi_c(\mathbb{R}^3)=-1$, $\mathbb{R}^3$ is not a topological square.

More generally, since $\chi_c(\mathbb{R}^n)=(-1)^n$, for any $k\in\mathbb{N}$, $\mathbb{R}^{2k+1}$ is not a topological square.

For an introduction to the topological Euler characteristic with compact support, I would recommend the following notes by LIVIU NICOLAESCU.

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    $\begingroup$ You say that $\chi_c$ is "available for any reasonable space" - does the argument above genuinely show that $\mathbb{R}^{2k+1}$ is not a topological square, or just that it isn't the square of a "reasonable" space in the appropriate sense? $\endgroup$ Oct 4, 2020 at 18:41
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    $\begingroup$ The answer assumes that the square root has a well defined euler characteristic with compact support. $\endgroup$
    – JME
    Oct 8, 2020 at 2:55

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