This is a question I have asked on mathstackexchange with a bounty but without any answer; it is probably more adapted to mathoverflow:
Let us assume that $\Sigma_n$ is a sequence of topological spheres into $\mathbb{R}^3$. We assume that the surfaces are Alexandrov embedded, that is to say there exist some immersion $i_n$ from $B^3$ into $\mathbb{R}^3$ such that $i_n(S^2)=\Sigma_n$.
Then we assume that $\Sigma_n$ converges to $\Sigma$ in the sense that $i_n:S^2 \rightarrow \mathbb{R}^3$ such that $i_n$ converges in $C^2(S^2)$ to $i:S^2 \rightarrow \mathbb{R}^3$.
Is that possible that $\Sigma$ gets a branch point, i.e. locally $i$ looks like $z↦z^k$? It is easy to prove if we assume that $\Sigma^n$ is embedded since we can't converge locally to some $z\rightarrow z^k$: just look to the winding number of the curve given by the intersection of the surface and a thin cylinder centred at the branched point.
This question comes from constant mean curvature surface theory in some pertubative setting. Hence I can assume that the branch points are isolated.
For instance: does there exists a sequence of smooth Alexandrov embedded surfaces which converge to $\omega \circ z\rightarrow z^3$ where $\omega$ is the inclusion of $S^2$ into $\mathbb{R}^3$?
