Prove $\mathbb{R}^3$ is not the product of two identical topological spaces

Solution 1:

As has been pointed out in the comments, this question has been asked and answered on MathOverflow. I have replicated the accepted answer by Tyler Lawson below.

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 $X \setminus \{p\} \subset X$ and a field $k$, we have isomorphisms (the relative Kunneth formula) $$ H_n(X \times X, X \times X \setminus \{(p,p)\}; k) \cong \bigoplus_{p+q=n} H_p(X,X \setminus \{p\};k) \otimes_k H_q(X, X \setminus \{p\};k). $$ 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 $H_p(X, X \setminus \{p\};k)$ were nontrivial in degree $n$, then the left-hand side must be nontrivial in degree $2n$.