Existence of a homeomorphism of an $n$-dimensional cube I am struggling to come up with a formal argument for why a homeomorphism with the following properties exists:
Let $n\in \mathbb{N}$, $t\in [-1, 1)$, $J = [-1, 1]$ and $\varepsilon>0$, there is a homeomorphism $h:J^{n+1}\to J^{n+1}$ such that $h$ is the identity function on $J^n\times [-1, t]$ and $\text{diam}(h(J^n \times\{1\})) < \varepsilon$.
For $n = 1$ I can see such homeomorphism exists using barycentric coordinates with triangles.
There might be a way to prove this by induction, but I don't see how.
 A: Here's a sketch of a construction. We want to map $J^{n} \times [t, 1]$ homeomorphically to itself so that the slice $J^{n} \times \{t\}$ is mapped by the identity and $J^{n} \times \{1\}$ is "squeezed" to arbitrarily small diameter.
Fix $\varepsilon > 0$ arbitrarily, set $t' = \frac{1}{2}(1 + t)$, let $r$ be the piecewise-linear function equal to $1$ for $t \leq u \leq t'$ and equal to $\varepsilon/(4\sqrt{n})$ at $u = 1$, and perform "axial affine scaling"
$$
(x, u) \mapsto (r(u)x, u),\quad x \in J^{n}, t \leq u \leq 1.
$$
The image of this continuous bijection (blue) is the block $J^{n} \times [t, t']$ surmounted by a truncated pyramid over $J^{n}$ comprising all $(x, u)$ such that $t' \leq u \leq h(x)$ for some continuous, piecewise-linear function $h:J^{n} \to [t', 1]$.
Next consider the $x$-dependent affine scaling
$$
(x, u) \mapsto (x, t + (1 - t)(u - t)/(h(x) - t)),\quad t \leq u \leq h(x),
$$
which stretches each vertical segment $x \times [t, h(x)]$ to $x \times [t, 1]$ continuously in $x$, giving a continuous bijection from the blue solid to $J^{n} \times [t, 1]$. The composition is a continuous bijection from $J^{n} \times [t, 1]$ to itself, and therefore a homeomorphism since $J^{n} \times [t, 1]$ is compact.

