This question arose from this stack overflow post. For a topological space $Y$, if there exists a topological space $X$ such that $X \times X$ is homeomorphic to $Y$ then $X$ is called a "topological square root" of $Y$. I was wondering now whether this topological square root is well defined up to homeomorphism, that is to say: \begin{equation} A \times A \cong B \times B \implies A \cong B \end{equation}
I am aware that the implication holds the other way, since if $f: A \to B$ is a homeomorphism then so is $f \times f : A \times A \to B \times B$, however I've not been able to prove whether the implication in the equation above is true and am now starting to suspect that there is indeed a counterexample.
I'm aware of products not being "cancellable", that is to say, if $M \times X \cong N \times X$ then it is not necessarily true that $M \cong N$, with a counterexample being $X = \mathbb{R}$, $N = \mathbb{R}^3$ and $M$ being the Dogbone Space. This suggests that perhaps there is a counterexample to the question posed, though I'm not sure since in this case there's only 2 spaces.
