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6 votes

For $\mathbb R^n \times Q \cong \mathbb R^m \times Q $ must $n = m$? ($Q$ is the Hilbert cube)

Note that $\mathbb{R}^n\times Q$ is proper homotopy equivalent to $\mathbb{R}^n$. If they were homeomorphic, we would obtain a proper homotopy equivalence between $\mathbb{R}^m$ and $\mathbb{R}^n$. Us …
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