Timeline for Universal semistable curve

4 events

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Dec 3 at 14:47	comment	added	Jason Starr		Oops, I was wrong. The image of the $1$-morphism is open. However, the original $1$-morphism is not immersive. Consider first one of the three stable $4$-pointed curves in the boundary of $\overline{\mathcal{M}}_{0,4}$. Then consider the associated semistable curve that has one (unstable) rational bridge between the two components of the stable curve. When we add a fifth marked point on the rational bridge, how do we know which of the two semistable curves it comes from? In fact, it comes from both.
Dec 3 at 13:32	comment	added	E. KOW		Thank you! Do you know a reference for the last statement (open immersion) by any chance?
Dec 3 at 11:59	comment	added	Jason Starr		The usual notation for the stack that you denote $\overline{\mathcal{M}}^{\text{ss}}_{g,n}$ is $\mathfrak{M}_{g,n}$. Yes, there is an $1$-morphism from the universal curve over $\mathfrak{M}_{g,n}$ to $\mathfrak{M}_{g,n+1}$ in the way that you describe. You do also need to address what happens when $x$ "crosses" one of the $n$ marked points $y_\ell$. In that case, you "sprout" a rational tail that is attached to the original curve at $y_\ell$ and contains $x$ as well as the $\ell^{\text{th}}$ marked point (and no other marked points). The $1$-morphism is representable by open immersions.
Dec 2 at 23:53	history	asked	E. KOW	CC BY-SA 4.0