Given a 2-disc embedded in $\Bbb R^4$, can I fit another 2-disc with the same boundary? I am given a 2-disc $D^2$  embedded into $\Bbb R^4$, that is, I have an injective continuous map $\phi:D^2\to\Bbb R^4$. I want to "double" this disc in the sense that I am looking for a second embedded disc $\smash{\psi:D^2\to\Bbb R^4}$ that agrees with $\phi$ on the boundary $\smash{\partial D^2 = S^1}$ but whose image is otherwise disjoint from the image of $\phi$.

Question: Can I always find such a second embedded disc?

If it helps, we can assume that $\phi$ is piece-wise linear, but then $\psi$ should be as well (in fact, Will's comment shows that we should probably work in a category that does not contain an equivalent of Alexander's horned sphere).
I also believe this is equivalent to asking whether every embedding of the northern hemisphere $\subset S^2$ extend to an embedding of the full sphere.

If $\phi$ were differentiable ...
... (at least in the interior of $D^2$) then I believe we can choose a continuously varying normal vector $n:\mathrm{int}(D^2)\to\Bbb R^4$ at each interior point of the disc and define
$$\psi(x):=\phi(x)+\epsilon(x)n(x),$$
where $\epsilon(x)$ is positive but sufficiently small on $\mathrm{int}(D^2)$ and tends to zero as $x$ approaches $\partial D^2$ (so I don't care what $n(x)$ is on the boundary).
But I do not want to assume differentiability and so I have no idea for how to choose the normal vector at each point.
 A: Without PL structure and smoothness, the counterexamples are constructed by Bob Daverman. That is, there exist wildly embedded 2-disks in $\mathbb{R}^n$ ($n\geq 4)$. In fact, those disks have non-simply connected complement in $\mathbb{R}^n$.  See

*

*On the absence of tame disks in certain wild cells. Geometric topology (Proc. Conf., Park City, Utah, 1974), pp. 142–155. Lecture Notes in Math., Vol. 438, Springer, Berlin, 1975. MR0400236

*On the scarcity of tame disks in certain wild cells.
Fund. Math. 79 (1973), no. 1, 63–77. MR0326742

A: Yes, let us assume that $D$ is PL.
Let us move all inner vertices of $D$ randomly a bit.
Denote by $D'$ the new disc; it might intersect $D$ at a collection of isolated points $p_1,\dots,p_n$; we can assume that no $p_i$ belongs to 1-skeleton.
For each $p_i$ choose a polygonal path $\gamma_i$ to $\partial D$ in $D$;
we may assume that $\gamma_i$ do not intersect each other and they do not pass thru the vertices.
Note that one can remove the intersection point $p_i$ by modifying $D'$ in a small neighborhood of $\gamma_i$ (a simplified version of the Whitney trick).
