Why is it true that a closed proper subvariety of a curve is a finite set of points?
I had the following lines of thinking:
Let $C$ be a curve and $X$ a proper closed subvariety of $C$. Write $C$ as a union of its irreducible components $C = C_0 \cup \ldots \cup C_r$. Then $ X = X_0 \cup \ldots \cup X_r$ where $X_i$ = $X \cap C_i$ and each $X_i$ is closed in $C_i$. Since $C$ is a curve, each of the $C_i$ have dimension $1$. So each $X_i$ must be a single point (is this true?) since a single point is always closed. Thus $X$ consists of $r+1$ points.
If my reasoning is correct, then great. The reason I think it isn't is because in my notes I have the following Lemma:
Let $\alpha : X \to Y$ be a non-constant morphism of irreducible curves. Then
i) $\forall y \in Y$, $\alpha^{-1}(y)$ is a finite set
ii) ...
Using the same reasoning as before, I'd show that $\alpha^{-1}(y)$ consists of just $1$ point; rather than "finitely many".
Thanks