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Let $S$ be a scheme. By a smooth curve of genus $g$ over $S$ we mean a proper, flat, family $C \to S$ whose geometric fibers are smooth, connected $1$-dimensional schemes of genus $g$. The moduli functor $\mathcal{M}_g$ of smooth curves of genus $g$ over a noetherian base $S$ is the functor that sends each $S$-scheme $B$ to the set $\mathcal{M}_g(B)$ of isomorphism classes of smooth and proper morphisms $C \to B$(where $C$ is also an $S$-scheme) whose fibers are geometrically connected curves of genus $g$.

I have a question about the argument that this moduli functor $\mathcal{M}_g$ of smooth curves of genus $g$ over a noetherian base $S$ is not representable (in spirit of Yoneda-Lemma this means that $\mathcal{M}_g$ isn't a sheaf). I found it in Pedro Castillejo's paper Introduction to stacks and he refered for a "detailed" proof to Dan Edidin's "Notes on the construction of the moduli space of curves". Now one argument from Edidin's paper I not understand:

The key point is that some curves have non trivial isomorphisms, i.e. not identity maps, and it makes it possible to construct non-trivial families $C \to B$ where each fiber has the same isomorphism class. The construction in Edindin's paper on page 3 I understand.

What I not understand is why the existence of such non-trivial family $C \to B$ of isomorphic fiber imply that the functor $\mathcal{M}_g$ is not representable. In the paper on page 2 the argument is

...As a result, it is possible to construct non-trivial families $C \to B$ where each fiber has the same isomorphism class. Since the image of $B$ under the corresponding map to the moduli space is a point (???), if the moduli space represented the functor $\mathcal{M}_g$ then $C \to B$ would be isomorphic to the trivial product family (???)

Assume $\mathcal{M}_g$ is representable by an $S$-scheme $M$, thus we have natural equivalence $\mathcal{M}_g \cong Hom_S(-, M)$.

Q_1: Why this imply that then the image of $B$ is a point of $M$?

Q_2: I not see how the sheaf axiom would give a contradiction to representability.

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Suppose $M$ represents your functor $\mathcal{M}_g$ and you have an isotrivial family $C\rightarrow B$ which is globally non-trivial. Then by definition, there exists a universal family $\mathcal{C}_g \rightarrow \mathcal{M}_g$ map $B\rightarrow M_g$ such that the pullback (fiber product) gives you $C\rightarrow B$.

Now if you take any $b\in B$, then under the composition $b\hookrightarrow B \rightarrow M_g$ corresponds to the family $C_b\rightarrow b$, where $C_b$ is the fiber of $C\rightarrow B$ over $b\in B$.

Since $C\rightarrow B$ is assumed to be isotrivial, $C_b \equiv C_{b'}$ for any $b,b' \in B$. Therefore $b,b'$ will map to the same point in $\mathcal{M}_g$. In particular we deduce that $B\rightarrow \mathcal{M}_g$ is the constant map, where every point maps to the point corresponding to the curve $C_b$.

However since $C\rightarrow B$ is the fiber product, it follows that $C\rightarrow B$ is the trivial family. This contradicts the assumption that $C\rightarrow B$ is non-trivial.

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  • $\begingroup$ $2\frac{1}{2}$ remarks on your answer: the "universal family" $\mathcal{C}_g \rightarrow \mathcal{M}_g$ is exactly that one that determines the natural isomorphism $\mathcal{M}_g \cong Hom(-,M)$ in spirit of Yoneda-Lemma as explaned here ncatlab.org/nlab/show/representable+functor,right? And by $M_g$ you mean $M$? $\endgroup$ – user526728 Dec 24 '19 at 6:24
  • $\begingroup$ Secondly, I understand how you conclude that $B\rightarrow \mathcal{M}_g$ must be a constant map with image $c \in M$ corresponding to curve $C_g$. What I not understand is why does this imply that $C \to B$ most be the trivial family? $\endgroup$ – user526728 Dec 24 '19 at 6:24
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    $\begingroup$ @TimGrosskreutz 1. Yes and yes -- oops! 2. So if you know that it's the constant map, then you know that the map $B\rightarrow M$ factors through some point $B\rightarrow b = \operatorname{Spec} k \hookrightarrow M$. The map $b\rightarrow M$ also corresponds to a family, and it's precisely the corresponding curve over a point. i.e. it's $C_b\rightarrow \operatorname{Spec} k$. Now you take the fibered product of $B\rightarrow b = \operatorname{Spec} k$ and $C_b \rightarrow \operatorname{Spec} k$ and you'll get a trivial family. $\endgroup$ – loch Dec 24 '19 at 7:26

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