The map $\operatorname{Log}|_{\mathbb{R}H^{\circ}}:\mathbb{R}H^{\circ}\to\mathcal{D}\subset\mathbb{R}^{n+1}$ is an immersion since the map $\operatorname{Log}|_{(\mathbb{R}^{*})^{n+1}}:(\mathbb{R}^{*})^{n+1}\to\mathbb{R}^{n+1}$ is an immersion (it is a trivial $2^{n+1}$-covering of $\mathbb{R}^{n+1}$).

To see the transversality we recall the definition of the foliation $\mathcal{F}^{\prime}$. For each component of $\mathbb{R}^{n+1}\smallsetminus\Sigma_{n}$ the foliation $\mathcal{F}^{\prime}$ is parallel to a vector $\stackrel{{\scriptstyle\to}}{{v}}$ normal to a facet of the Newton polyhedron of $H^{\circ}$. Therefore, any hyperplane in the image $\gamma(\mathbb{R}H^{\circ})$ is transverse to $\stackrel{{\scriptstyle\to}}{{v}}$. Furthermore, hyperplanes close to being parallel to $\stackrel{{\scriptstyle\to}}{{v}}$ are close to the hyperplane in ${\mathbb{C}}{\mathbb{P}}^{n+1}$ corresponding to this facet and therefore are far from the given component of $\mathbb{R}^{n+1}\smallsetminus\Sigma_{n}$. Thus the result $\mathcal{F}$ of smoothing is also transverse to $\mathcal{D}$ and the angle between them in $\mathbb{R}^{n+1}$ is separated from 0. ∎