For the moduli of stable curve, we have the result by Deligne-Mumford-Knudsen that there's an isomorphism of moduli spaces
$$\Phi : \overline{\mathcal{X}}_{g,n} \xrightarrow{\sim} \overline{\mathcal{M}}_{g,n+1}$$
over $\overline{\mathcal{M}}_{g,n}$, where $\overline{\mathcal{X}}_{g,n}$ is the universal curve over $\overline{\mathcal{M}}_{g,n}$ and the map is given by stable reduction.
My question is what happens in the semistable case: As was mentioned in this MO post, the moduli of semistable curves is algebaic. I believe semistable reduction gives a map $$\Phi^{\operatorname{ss}} : \overline{\mathcal{X}}^{\operatorname{ss}}_{g,n} \to \overline{\mathcal{M}}^{\operatorname{ss}}_{g,n+1}$$ Take $g>0,n=0$. A point $(X,x)$ of $\overline{\mathcal{X}}^{\operatorname{ss}}_{g,0}$ will fail to be in $\overline{\mathcal{M}}^{\operatorname{ss}}_{g,1}$ only if $x$ is not smooth, in which case the semistable reduction will be the curve $X$ with a rational bridge added at $x$ and with the point $1$ marked. On the other hand for a smooth point $y \in X$, the $1$-pointed semistable curve $(X \vee_{y \sim 0} \mathbb{P}^1,\infty)$ is not of this form. So it seems $\Phi^{\operatorname{ss}}$ is not surjective. Is it an embedding?
