Morphism to moduli stack of curves

Question

I am currently reading up on moduli stack of algebraic curves (https://www.math.uni-bonn.de/~schmitt/ModCurves/Script.pdf). On Page 54 in this script, they claim that giving a morphism $f: S \to \mathcal{M_{g,n}}$ is equivalent to giving a family $\pi: C\to S$ of curves over $S$.

However, I don't know how to show this equivalence and don't even know how to start a proof.

Any help is appreciated :)

Answer 1

This is essentially the Yoneda lemma. Every category $\pi\colon\mathcal{M}\to\mathbf{Sch}$ fibered in groupoids is equivalently described as a $(2,1)$-functor $F_\mathcal{M}\colon\mathbf{Sch}^\mathrm{op}\to\mathbf{Grpd}, X\mapsto \pi^{-1}(X)$, where $\mathbf{Grpd}$ is the $(2,1)$-category of groupoids. This is the Grothendieck construction, or straightenening-unstraightening if you're a fan of higher categories: there is an equivalence of $(2,1)$-categories between the $(2,1)$-category of categories $\mathcal{M}\to\mathbf{Sch}$ fibered in groupoids and the $(2,1)$-category of functors $\mathbf{Sch}^\mathrm{op}\to\mathbf{Grpd}$. See for instance Proposition 3.3 here: https://ncatlab.org/nlab/show/Grothendieck+construction#statement_of_the_equivalence .

Therefore, a morphism $S\to\mathcal{M}$ of categories fibered over groupoids is the same data as a natural transformation $yS\to F_\mathcal{M}$ of functors $\mathbf{Sch}^\mathrm{op}\to\mathbf{Grpd}$, where $yS$ denotes the image of $S\in\mathbf{Sch}$ under the Yoneda embedding, followed by the inclusion $\mathbf{Set}\hookrightarrow\mathbf{Grpd}$ (this is also how the $(2,1)$-categorical Yoneda embedding acts on $S$). Now we invoke the $(2,1)$-categorical Yoneda lemma, which states that there is now an equivalence of groupoids between such natural transformations $yS\to F_\mathcal{M}$ and $F_\mathcal{M}(S)\simeq \pi^{-1}(S)$.

Answer 2

If you really have no idea where to start, then I'm not sure these lecture notes are the best thing to be reading.

That said, a moduli space $\mathcal M$ should, essentially by definition, carry a bundle $u\colon \mathcal E\to \mathcal M$ over it which has fibre $u^{-1}(s)$ over $s$ equal to the object it represents (in the case of $\mathcal M_{g,n}$ therefore, a curve). Given any map $f\colon S \to \mathcal M$, you can then pull this bundle back to $S$ to get $S\times_f\mathcal E$, a bundle, or family, over $S$.

To take a more concrete example though, if $\mathcal M = \mathbb P^1$, then a point in $\mathcal M$ is a line in $\mathbb C^2$, so that $\mathbb P^1$ is the moduli space of lines in the plane, and these lines collectively give you the tautological line bundle $\pi \colon \mathcal L \to \mathbb P^1$ over $\mathbb P^1$. A map $f\colon S \to \mathbb P^1$ thus gives you a family over $S$ of lines in $\mathbb C^2$, where the line over $s\in S$ is just $f(s)$, or in other words, $f$ lets you pullback $\mathcal L$ so that you obtain a line bundle $f^*(\mathcal L)$ over $S$.

The fact that this gives an equivalence is simply saying that the family over $S$ gives you a map to $\mathcal M$ because if $s\in S$ then $f^{-1}(s)$ can be viewed as a point in the moduli space $\mathcal M$.