Moduli functor $\mathcal{M}_g$ of smooth curves of genus $g$ not representable 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.
 A: 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. 
