Is morphism between curves projective? Suppose $X$ is a smooth, projective curve, $Y$ is an arbitrary curve(may be singular), and both curves are over an algebraically closed field $k$ with character 0. Let $f: X \to Y$ be a morphism between curves. Is $f$ a projective morphism? Here, projective morphism is in the sense of Hartshorne, i.e. $X \to Y$ factors through $X \to \mathbb{P}^{n}_{Y} \to Y$, with $X \to \mathbb{P}^{n}_{Y}$ a closed embedding, $\mathbb{P}^{n}_{Y} \to Y$ the the natural projection to $Y$ factor.
I guess it is projective by the following general heuristic argument:
Statement:Suppose $X \subset \mathbb{P}^{n}$ is a closed subvariety, $Y$ is another variety, then any morphism $f: X \to Y$ is projective.
One can define $f' :X \to \mathbb{P}^{n}_{Y}$ by $x \mapsto (x,f(x))$, and this is an injective map. Moreover, because $X$ is proper, its image must be closed. I guess these guarantee $f'$ is a closed embedding, and the projection $\mathbb{P}^{n}_{Y} \to Y$ is easy to define.
I am not quite sure about the above argument, especially $f'$ being a closed embedding. Any suggestions? 
 A: The answer is yes. Your proof looks fine to me. If you want a way to do it with less explicit computation, are you familiar with the notion of a base change? If you are, The map you define is the composition of $X \to X \times Y$ and $X \times Y \to \mathbb{P}^n \times Y$. The first one is a base change from $Y \to Y \times Y$, and is thus a closed embedding. The second one is a base change of $X \to \mathbb{P}^n$. This proves that $f′$ is a closed embedding.
There is also a more general statement (Which I know as the cancellation theorem) that is very useful: Let $X, Y$ be $Z$-schemes, with structure morphisms $g$ and $h$, and let $f:X \to Y$ be a map of $Z$-schemes. Then f is just the composition of the graph morphism $X \to X \times_Z Y$ and the projection $X \times_Z Y \to Y$. The most important application of this is that if we have some property P of morphisms closed under composition/base change, then to prove $f$ is P, it suffices to show $g$ and the diagonal morphism of h have property P.
