Given a morphism $\phi : X\rightarrow Y$ of smooth projective varieties and a non-constant
map $f : C\rightarrow Y$, where $C$ is amooth projective curve, how to construct a smooth
projective curve $C'$ with non-constant map $g : C'\rightarrow X$ such that $\phi \circ g = f\circ h$, where $h : C'\rightarrow C$.



smooth projective curve
First form the fibre product $Z$ of $X$ and $C$ over $Y$, to get maps $f':Z \to X$ and 
$\phi':Z \to C$ such
that $\phi\circ f' = f \circ \phi'.$


Now if the map $\phi'$ is not surjective (equivalently, dominant, since all varieties in sight are projective) then we can't find $C'$. (This happens when the image of $f$ is not contained in the image of $\phi$.) [EDIT: As Sandor Kovacs and Roy Smith noted, I am taking it as a condition that $h$ is also non-constant, although this is not explicitly stated in the question; otherwise one could allow more degenerate solutions. Also, I am assuming that all curves under discussion are connected.]


Suppose now that $\phi'$ *is* surjective. Then we can find a one-dimensional subvariety
$D$ of $Z$ such that $\phi'(D) = C$. Indeed, since $\phi'$ is surjective, we may find at least one component of $Z$ such that the restriction of $\phi'$ to that component is already surjective. Replace $Z$ by this component, and so assume that $Z$ is irreducible.
If $Z$ is also one-dimensional, then we can take $D = Z$. If not, a general hyperplane section of $Z$ will also surject onto $C$, and so we may replace $Z$ by this hyperplane section, thus reducing the dimension. Continuing in this way, we eventually find
the desired $D$.


Finally, we can form the normalization $C'$ of $D$. The composite 
$$C' \longrightarrow D \hookrightarrow Z \buildrel f' \over \longrightarrow X$$
is then the desired map $g$, while the composite
$$C' \longrightarrow D \hookrightarrow Z \buildrel \phi' \over \longrightarrow C$$ 
is the desired map $h$. 


