The definition below is from Encyclopaedia of Mathematics: Volume 6.
Question: For any $n\geq 1$, is the $n$-dimensional closed cube $$[0,1]^n=[0,1]\times [0,1]\times\cdots \times[0,1]$$ isotopy equivalent to the $n$-dimensional open cube $$ (0,1)^n=(0,1)\times (0,1)\times\cdots\times (0,1)? $$ I guess it is true but do not know how to prove...
I am not sure If I understand the definition correctly, but if I do, then take a look at the maps $$ f : (0,1) \rightarrow [0,1], x \mapsto x$$ and $$ g : [0,1] \to (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$ These are embeddings and composing these we get $$ f \circ g : [0,1] \rightarrow [0,1], x \mapsto \frac{1}{3} x + \frac{1}{3}$$ and $$ g \circ f : (0,1) \rightarrow (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$ Convex combination with the identity gives us $F_t : [0,1] \rightarrow [0,1]$ defined by $$F_t(x) = \left(\frac{1}{3}t + (1-t)\right) x + \frac{1}{3}t$$ which is an isotopy with $F_0$ being the identity map and $F_1 = f \circ g$. The same map restricted to $(0,1)$ also yields an isotopy between the identity and $g \circ f$. If my understanding is correct, this settles the case $n = 1$ and for arbitrary $n$ we can use $f$ and $g$ in every component.