TITLE: Is $\mathbb R^3$ the square of some topological space?
QUESTION [213 upvotes]: 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 $\mathbb{R}^3$?

REPLY [17 votes]: 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.