Let $D_1$ and $D_2$ be two open disks embedded in $\mathbb{R}^n$ such that we have an inclusion $D_1 \hookrightarrow D_2$. Both $D_1 \stackrel{f_1}{\cong} D $ and $D_2 \stackrel{f_2}{\cong} D$ are diffeomorphic to the same open disk $D$. Is it then possible to produce an isotopy between the inclusion and the composition $f_2^{-1}\circ f_1$?
The only way I know how to produce isotopies is by a linear function, but in this situation this will not always give an embedding for each $t\in [0,1]$ (I have a simple illustration for $n=2$ I can include if it's helpful).
Any pointers to what I could try/where to look are highly appreciated!
Yes, you can construct an isotopy with Alexander's trick by adding as hypothesis $f_2^{-1}\circ{f_1}|_{\partial{D_1}}=i|_{\partial{D_1}}$.
$\textbf{Alexander's Trick.}$ Let $f,g : D^n \longrightarrow{D^n}$ be two homeomorphism between the $n$-dimensional closed disk in $\mathbb{R}^n$, such that $f(x)=g(x)$ for all $x\in{\partial}D^n=S^{n-1}$, then $f$ is isotopic to $g$.
In particular, an isotopy between $f$ and $id_{D^n}$, such that $f|_{\partial{D^n}}=id_{\partial{D^n}}$, is given by $$H:D^n\times[0,1]\longrightarrow{D}^n\qquad H(x,t)= \begin{cases} t{\cdot}f(\frac{x}{t}) & 0\leq||x||<t\\ x & t\leq||x||\leq1 \end{cases}$$