Alexandrov embedded disc We say that a compact surface $\Sigma$ is Alexandrov embedded via an immersion $f:\Sigma \rightarrow \mathbb{R}^3$ if there exist, $X$ a 3 manifold and an immersion $F: X \rightarrow  \mathbb{R}^3$ such that $\partial X =\Sigma$ and $F_{\vert \Sigma}=f$.
The Wente Torus is an example of an immersion which is not Alexandrov embedded.
My question first is probably simple for a topologist: does every immersion of $S^2$ is Alexandrov embedded? I am pretty sure that the answer si yes, thanks to the Smale classification of sphere immersion. But I look for an easy argument.
Does this notion can make sense for a disk immersion:
1st idea: if we fix the boundary, I think there is easy example of disc immersion which can not be extended to a half ball, but I am not sure.
2nd idea: is every gluing of a second disc to the first one on the boundary is equivalent? In that case we are back to my question about sphere.
 A: Claim. There exist immersions $f: S^2\to R^3$ which do not extend to immersions $X\to R^3$ for arbitrary compact connected 3-dimensional manifold $X$ with boundary $S^2$. 
As a concrete example, consider an immersion $h: S^2\to R^3$ such that $h$ has only double self-intersection points, namely, there exist two disjoint smooth circles $C_1, C_2\subset S^2$ with $h(C_1)=h(C_2)=C$ and $h$ is 1-1 on the complement to the union of these circles. The complement $R^3\setminus S=h(S^2)$ consists of three components: One unbounded component $U$ and two bounded components $V$ and $W$. 
This immersion, of course, extends to an immersion $\tilde{h}: B^3\to R^3$ so that $\tilde{h}$ is 1-1 over $V$ and 2-1 over $W$. (A picture would be very helpful here, but I am not good with pictures.) 
Now, consider $S^3$ as the 1-point compactification of $R^3$ and let $p$ be a point in $V$. We then take $J$ to be the inversion sending $p$ to $\infty$ and let $f=J\circ h$. I claim that $f$ is our map. Indeed, suppose that $h: S^2\to S^3$ extends to an immersion $g: X\to S^3$ whose image misses $p$ (where $X$ is a compact manifold with boundary $S^2$). Then the image of $g$ is also disjoint from the entire $V$. 
Of course, the image of $g$ has to intersect both $U$ and $W$. However, then, the preimage of $C$ under $g$ disconnects $X$ (since $C$ disconnects $\bar{U}\cup \bar{W}$). Since $g$ is an immersion, $g^{-1}(C)$ is 1-dimensional. Contradiction. 
Note that the same proof would have worked for $g$ which is locally a branched cover. 
One can use a similar argument to show that if $h: S^2\to R^3$ is a generic immersion which is not 1-1, then composition $f=J\circ h$ (with a suitably chosen inversion $J$) does not extend to an immersion $X\to R^3$. 
One the other hand, one can use Alexander's arguments (from his proof that every oriented n-manifold is a branched covering over the n-sphere) to prove the following:
Theorem. Let $S$ be a smooth closed surface, $X$ is a compact smooth oriented 3-dimensional manifold with boundary $S$. Suppose that $f: S\to S^3$ is an immersion. Then $f$ extends to a smooth map $\tilde{f}: X\to S^3$ which is locally a branched covering. 
