I'm reading the Otal's survey on the compactification of Morgan Shalen. (available here)

He claims in an example (page 8) that the compactification of $\mathbb C^2$ is $S^4$, which sounds completely natural, but I'm not able to understand why, and if I do the calculations I obtain a different object.

Here the definition:

Let $X$ be an affine algebraic set and take a finite (or countable) generator set $F$ of the ring of the regular functions on $X$. In the example I'm interested in $X=\mathbb C^2$ and the coordinates $z,w$ are my generating set.

Let $[0,\infty)^F$ and denote by $\mathbb P^F$ its projectivized, with $\pi$ the natural projection. In my example, $F$ has two elements so $\mathbb P^F$ is a closed segment.

Define $\theta_0:X\to [0,\infty)^F$ by $\theta_0(x)=(\log(|f(x)|+2))_{f\in F}$ and

$\theta=\pi\circ\theta_0$.

Let $\hat X$ be the one-point compactification of $X$.

The MS compactification of $X$, w.r.t. $F$ is the closure of the graphic of $\theta(X)$ in $\hat X\times\mathbb P^F$

In the example the function $\theta$ is $$\theta(z,w)=\dfrac{\log(|z|+2)}{\log(|w|+2)}$$

By studying the level sets of such function it seems to me that the compactification of $\mathbb C^2$ is a singular object, while in the survey is claimed that it is $S^4$.

In particular the closure of a level set in $\hat X\times \mathbb P^F$ is the one-point compactification of the level set itself. Level sets are $|z|+2=(|w|+2)^c$ so at infinity they are of the form $T^2\times[a,\infty)$ whose one-point compactification is the cone over $T^2$, which is singular. As level sets are disjoint this would show that the compactification of $\mathbb C^2$ is not a manifold.

My question is: Is there any place where i can find the proof that the MS compactification of $\mathbb C^2$ is $S^4$? Or, can anyone give some hint?