My intuition tells me that any two topological embeddings of the closed 2-ball (aka unit disk) into $R^3$ are isotopic in $R^3$. Is this correct? Maybe easy to prove?

It seems like it should be easy because the disk is compact and contractible, but I sill don't know what to do about it. I am working on problems in mathematical origami and I'm hoping to use this fact as a starting point in a much bigger proof.

This is not my area of expertise, so any tips on what to search or how to go about proving (or disproving) this would be helpful. I have searched different ways of wording this question and haven't found anything helpful.

I suspect this may generalize to closed $n$-balls embedded in $R^{n+1}$ or even an $(n+1)$-manifold, but I only need the case $n=2$.

Update: After researching some more based on ARupinski's comment, I realized that I am concerned specifically with piecewise-linear embeddings of the disk into $R^3$ with finitely many pieces. I don't think you can make a piecewise-linear Alexander Horned Shere with only finitely many pieces. So my question would be, if $f$ is a piecewise-linear embedding of the closed 2-ball $B^2$ into $R^3$ with finitely many pieces, and $g$ is another embedding of $B^2$ into $R^3$ such that $g[B^2]$ is a closed triangle (with interior), is there an isotopy between $f$ and $g$ using piecewise-linear embeddings along the way? This sounds a bit easier to go about proving... if it's true... :( This is somewhat related to things like the Carpenter's Rule Theorem for chain linkages in the plane.