This question comes from the proof of "bend and break" lemma in "Higer-dimensional algebraic geometry" (p.59-60). I use the notations in compatible with the notation given there for convenience.
Let $X$ be a smooth projective variety, and let $C$ be a smooth curve over an algebraically closed field of character 0. Suppose there is a nonconstant morphism $f: C \to X$, and we choose a fixed point $c \in C$.
Let $ Mor(C, X\mid g:c\mapsto g(c))$ be the moduli space of morphisms from $C \to X$ with $c$ has fixed image $f(c)$, and let $T \subset Mor(C, X\mid g:c\mapsto g(c))$be a one dimensional subvariety (suppose it exists). We use $\bar{T}$ to denote a smooth compatification of $T$.
Because $T$ is a subvariety of moduli space, there is an evaluation morphism, and hence we have a map $$ev: C \times \bar{T} -\to C \times T \to X.$$ Suppose we can extend this map $ev: C \times \bar{T} -\to X$ to be a morphism, we wish to get a contradiction in the following setting:
(1) $C$ is not $\mathbb{P}^1$.
(2) The image of ev is in the curve $f(C)$, i.e. ev($C \times \bar{T}$)$=f(C)$.
According to the argument in the proof, because there is only finite automorprisms of $C$ (elliptic curve by Hartshorne IV Cor.4.7, higher genus by Hurwitz formula), there are only finite $t \in \bar{T}$, such that $ev(C,t) = f(C)$, and hence contradict the fact $\bar{T}$ is infinite.
However, I don't see how to show that "there are only finite $t \in \bar{T}$, such that $ev(C,t) = f(C)$".