A question on the morphism of projective varieties The continuation of this, my question


I want to show that $X$ and $Y$ are smooth and irreducible curves then $f(X)$ is either $Y$ or a point. 



Note that I know the proof of this theorem that the image of projective varieties under a regular map is closed. 
But I cannot apply this proof into my question. Please show me the proof of my statement. thank you. 
 A: First we have a theorem: If $X$ is an irreducible curve (affine or projective) then the only proper closed subsets of $X$ are finite sets of points. This requires a bit of machinery to prove, the proof that I know uses both nullstellensatz and transcendence dimension as a characterisation of dimension of varieties, and I don't think is particularly enlightening so I will leave it out. However, if you are willing to assume this, then we can now prove the statement:

Any morphism of irreducible projective curves $\alpha:X\rightarrow Y$ has image either a point, or $Y$.

We use the fact that given any non-constant morphism of irreducible projective curves $\alpha:X\rightarrow Y$, for any point $y \in Y, \#\alpha^{-1}(y)$ is finite. This follows from the first theorem since if $\alpha$ is non-constant, $\alpha^{-1}(y)$ is a proper closed subset of $X$ so is a finite set of points. 
Now, since $X$ has dimension $1$, $X$ is not a finite set of points, so the image of $\alpha$ cannot be finite. Using the above theorem again, since the image is closed in $Y$ and not a finite set of points, it must be all of $Y$. In other words, if $\alpha$ is non-constant, it is surjective.
