Identifying all but one face of $[0,1]^n$ Suppose $D=[0,1]^n, n \ge 2$. Then the quotient space $\frac{D}{\delta(D)\setminus [0,1]^{n-1}\times\{0\}}$ i.e. all but the bottom faces.
I could prove when $n=2$. In this case, we identify the three sides of the unit square. First, we take
$f: [0,1]^2\rightarrow S^1\times[0,1]$ by $f(x,y)=(e^x,y)$ If and after quotienting, we have $\frac{[0,1]^2}{\{(0,y),(1,y):y\in[0,1]\}}$ is homeo to $S^1\times [0,1]$. Further quotienting we get that the space we are looking for is homeo to reduced cone of $S^1$ which is actually homeo to the disc.
But how can we generalize it further? Are there any other ways to prove the same? I can easily visualize up to n=3 what might happen but can't produce an explicit map as in the case of n=2.
 A: The resulting space will always be homeomorphic to the $n$-dimensional disk $D = [0,1]^n$. I guess writing explicit formulas becomes a little annoying, but it is fairly clear what should happen if you divide the problem into two steps:

*

*First convince yourself that the cone $\operatorname{cone}([0,1]^{n-1}) = D/[0,1]^{n-1} \times \{1\}$ of the $(n-1)$-disk is homeomorphic to the $n$-disk $D$.

*Then convince yourself that there is a homeomorphism
\begin{aligned}
\phi: \frac{D}{\partial(D)\setminus [0,1]^{n-1}\times\{0\}} \xrightarrow{\;\;\;\cong \;\;\;} \frac{D}{[0,1]^{n-1} \times \{1\}}.
\end{aligned}
One could define $\phi$ for example by letting it be piecewise linear, determined by

*

*letting $\phi(1/2,1/2,\dots,1/2,t) = (1/2,1/2,\dots,1/2,t)$ for all $t \in [0,1]$,

*sending $\partial [0,1]^{n-1} \times \{0\}$ homeomorphically to $\partial [0,1]^{n-1} \times \{1\}$,

*sending $\partial [\frac{1}{3},\frac{2}{3}]^{n-1} \times \{0\}$ homeomorphically to $\partial [0,1]^{n-1} \times \{0\}$,

*sending $\partial [0,1]^{n-1} \times \{1\}$ homeomorphically to $\partial [\frac{1}{3},\frac{2}{3}]^{n-1} \times \{1\}$.

