Is $\mathbb R^3$ the square of some topological space? 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 $\mathbb{R}^3$?
 A: 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.
A: 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 $2n$.
A: I didn't know that, but I did know this: we cannot have $S^2 = S\times S$ for any topological space $S$.
A: 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.
