Is the contraction morphism proper?

Question

This is quite possibly a stupid question, but it is pretty far from what I normally do, so I wouldn't even know where to look it up.

Let $X$ be a projective variety and $F\in \overline{NE}(X)$ an extremal face. A morphism $f:X\longrightarrow Y$ is called the contraction of $F$ if the following hold:

1. $f(C)=$point for an irreducible curve $C\subset X$ iff $[C]\in F$,
2. $f_*(\mathcal{O}_X)=\mathcal{O}_Z$.

Then is the contraction $f$ proper?

Answer 1

If $X$ is proper, any morphism from $X$ is proper. So yes.

Edit: Since the answer is more or less trivial, I thought I'd add a bit on the intuition for proper morphisms $f : X \to Y$ in my head: Take a tiny curve $C$ downstairs in $Y$, and lift all but one point $c \in C$ to a (punctured) curve $\tilde C \subset X$ (so $f(\tilde C) = C - \{c\}$). Does there exist a way to "take the limit" for $\tilde C$, filling in the missing point upstairs? If the answer is always yes, the map is proper. (With this criteria it's easy to understand why $\mathbb{A}^1 \hookrightarrow \mathbb{P}^1$ isn't proper, for example.) But in this case $X$ itself is proper (i.e. compact), meaning we can just always take limits in $X$, for arbitrary curves.

Edit2: As pointed out by Hoot in the comments, I neglected to mention that the above intuition is precisely captured by the valuative criterion for properness.