Consider a proper flat morphism of $k$-schemes ($k$ is an algebraically closed field) $ f:X\longrightarrow\mathbb P^1\_k$ such that every fiber $X\_p$ for $p\in\mathbb P^1\_{\mathbb C}$ is a reduced connected projective *stable* curve of genus $g$. 



> 
> $X$ is said an **isotrivial family of curves** if there is an open
>  dense subset $U\subseteq\mathbb P^1\_{\mathbb C}$ such that all fibers
>  $\{X\_p\,:\, p\in U\}$ are smooth and isomorphic.
> 
> 
> 


In literature one can find also the following alternative definition of isotriviality:



> 
> $X$ is isotrivial if the the modular map $\varphi\_f:\mathbb
>  P^1\_{\mathbb C}\longrightarrow \overline{M\_g}$ is constant. (Here $\overline{M\_g}$ is the moduli space of stable curves).
> 
> 
> 


I don't understand why the two definitions are equivalent: in particular the first definition says that the moduli morphism is constant on the open dense subset $U$, but why can we conlude that it is constant on all $\mathbb P^1\_{\mathbb C}$ ($\overline{M\_g}$ is not Hausdorff)?



Isotriviality: two definitions
Let $f:X\to B$ be a stable curve of genus $g$ over a finite type integral $\mathbb C$-scheme.


Lemma 1. The moduli map $B\to \bar {M\_g}$ is constant if and only if there exists a dense open $U$ of $B$ such that all fibers of $X$ over $U$ are isomorphic.


Proof. (Compare with jmc's comment.) If $B\to \bar{M\_g}$ is constant, then the fibers of $X\to B$ are all isomorphic (by definition). So take $U=B$. Conversely, if all fibers over (some dense open) $U$ are isomorphic, then $U\to \bar {M\_g}$ is constant. The extension of the moduli map to $B$ (which exists by assumption) is then also constant. QED


Now, to answer your question, as Laurent Moret-Bailly points out, you need some regularity hypothesis as well.


Theorem. Suppose that $X$ is non-singular. Then the moduli map $B\to \bar {M\_g}$ is constant if and only if there exists a dense open $U$ of $B$ such that all fibers of $X$ over $U$ are **smooth** and isomorphic.


Proof. Since $X$ is nonsingular, the morphism $X\to B$ is smooth over some open $U$. In particular, if $B\to \bar{M\_g}$ is constant, then its image lies in $M\_g$. Conversely, apply the lemma with $B$ replaced by $U$. QED


