Proving that $(0,1)^3$ not homeomorphic to $[0,3)^3$

Question

What are some of the various ways of proving that $(0,1)^3$ is not homeomorphic to $[0,3)^3$ using the fundamental group and homology groups? I feel like I have various ways of understanding why this is true in terms of point-set topology but I would like some help with developing an algebraic approach.

Both spaces are contractible, so the fundamental group won't be able to tell them apart. Maybe if i remove a certain point from both of them it will? For that matter, is it wrong for me to think of $[0,1)^3$ as a closed octant of $\mathbb{R}^3$? That's what it looks like to me, as I believe $[0,1) \cong [0, \infty)$.

Thanks in advance!

Answer 1

Hint: Note that $[0,3)^3 \setminus \{ (0,0,0) \}$ is contractible, but $(0,1)^3 \setminus \{ x_0 \}$ is not contractible for any $x_0 \in (0,1)^3$.

Answer 2

$(0,1)^3$ is homogeneous and $[0,1)^3$ is not, as a point on the "boundary" (like $(0,0,0)$) cannot be mapped by a homeomorphism to a point of the interior, or vice versa. It probably can be reduced to an argument involving Brouwer's fixed point theorem e.g.