smooth projective curve 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 $h : C'\rightarrow C$.
 A: Assuming $\phi$ surjective, Matt's answer is OK if the ground field $k$ is perfect. Otherwise, one should weaken the conditions on $C'$ from "smooth" to "normal". 
For instance, assume $p=\mathrm{char}(k)>0$ and $a\in k$ is not a $p$-th power. Take $\phi:\mathbb{P}^2\to\mathbb{P}^2$ to be the Frobenius map $[x:y:z]\mapsto[u:v:w]=[x^p:y^p:z^p]$, and $C\subset\mathbb{P}^2$ to be the line $u+av=0$. Then $\phi^{-1}(C)$ is reduced and geometrically irreducible and  its function field is inseparable over $k$, so there is no nonconstant map from any smooth $k$-curve to $\phi^{-1}(C)$.
On the other hand, it's enough to assume $X$ and $Y$ proper and $\phi$ surjective (no smoothness or projectivity needed): let $z$ be the generic point of $C$, then $\phi^{-1}(z)$ is a nonempty $\kappa(z)$-scheme of finite type, hence has a closed point $z'$. The residue field of $z'$ is a one-variable function field over $k$, hence is the generic point of a unique normal projective irreducible $k$-curve $C'$. By the valuative criterion of properness, the canonical map $z'\to X$ extends to $C'\to Y$, and $C'$ is smooth if $k$ is perfect.
A: I am probably confused.  Matt's h seems always to be surjective.  What if C = Y, f = id, and C = X, and phi = constant map to point p of C?  Then we seem to need h to be constant. E.g. taking h = phi, and g = id, seems to work (and the minimum degree which occurs here is zero).  Is it possible your construction (essentially) works as long as the images of f and phi meet? I.e. that it works more widely, but that it is only when the image of f is actually contained in the image of phi that h is surjective?  No I guess you need the inverse image of f(C) in X, i.e. the fiber product, to contain a curve, in order for a non constant g to exist.  Something like that?
Indeed it seems more complicated than that.  What if we take f the inclusion of a plane curve C into P^2 = Y, and X the blow up of P^2 at a point p of C.  Then the we seem to be able to find both constant and non constant models for h.  I.e. C' = C, and either h = id, and g is the "proper transform" of f,  or h is a constant map from C to the point p, and g is any non constant map to the exceptional P^1 in X.
So there seem to be various answers of various degrees corresponding to different positive dimensional components of the fiber product, and whether or not they surject onto f(C).  Indeed if there is a single point f(p) on f(C), whose inverse image in X contains a curve D, then we seem to be able to take C' = D, and g= id, and h = the constant map to p.  Thus in general the minimum degree of h seems to be zero, e.g. whenever dimX > dimY and phi surjects.  Does this seem right?
A: 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 $h$.  
