Let $X$ and $Y$ be topological spaces. A homotopy from a continuous function $f: X \to Y$ to a continuous function $g: X \to Y$ is a continuous function $H : X \times [0, 1] \to Y$ such that $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$ for all $x \in X$. An isotopy is a homotopy such that $H(\cdot, t)$ is an embedding for every $t \in [0, 1]$; isotopy requires $f$ and $g$ to be embeddings.
Suppose $X = \{x \in \mathbb{R}^n : \sum_{i = 1}^n x_i^2 < 1\}$ is the open $n$-ball, and let $f, g : X \to Y$ be homeomorphisms. Does there always exist an isotopy from $f$ to $g$?
Alexander's trick solves a similar problem, but on closed $n$-balls.
Best Answer
Yes, but it is not as elementary as the Alexander trick. Your group is the same as the space of homeomorphisms of $\Bbb R^n$, and you want to know that the subspace of orientation-preserving homeomorphisms is connected. The case $n=1$ is easy to prove by hand, so we will suppose $n>1$.
In fact, $\text{Homeo}(\Bbb R^n) \cong \text{Homeo}(S^n, \text{pt})$, the space of homeomorphisms of the sphere that fix a point. This sits into a fibration $$\text{Homeo}(S^n, \text{pt}) \to \text{Homeo}(S^n) \to S^n,$$ and similarly for the orientation-preserving versions (which I denote $\text{Homeo}^+$).
So if $\text{Homeo}^+(S^n)$ is connected, we see from the long exact sequence $$\cdots \to \pi_1 S^n \to \pi_0 \text{Homeo}^+(S^n, \text{pt}) \to \pi_0 \text{Homeo}^+(S^n) \to \pi_0 S^n$$ that $\pi_0 \text{Homeo}^+(\Bbb R^n)$ is connected.
What remains is an inductive argument using the annulus theorem to show that this group is connected; I have written a proof out in the second half of this answer. The idea is to first demand your homeomorphism sends the equator to itself, inductively isotope that to the identity, and then use the Alexander trick on the two hemispheres.