TITLE: Given a 2-disc embedded in $\Bbb R^4$, can I fit another 2-disc with the same boundary?
QUESTION [11 upvotes]: 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.

REPLY [6 votes]: 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).

REPLY [6 votes]: 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