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 $\mathbb{R}^3$?



Is $\mathbb R^3$ the square of some topological space?
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 $2n$.


