TITLE: Characterization of a closed mapping
QUESTION [6 upvotes]: I'm having troubles with this basic question. I think it is really easy but I feel like blocked or blind with it.

Suppose we have a mapping $f\colon X \to Y$, the following are equivalent:
a) $f$ is closed,
b) $\forall U$ open on $X$, $\{y\in Y\mid f^{-1}(y)\subset U\}$ is open on $Y$,
c) $\forall C$ closed on $X$, $\{y\in Y\mid f^{-1}(y)\cap C\not= \emptyset \}$ is closed on $Y$.

I've tried seeing seeing that, as f is closed, and $X\setminus U$ is closed as $U$ is closed, then it's image is closed, so it complementary is open, and then try to see that $\{y\in Y\mid f^{-1}(y)\subset U\} = Y\setminus f(X\setminus U)$, but without success. And also try with interior and adherences, but not success again.
I hope somebody can give me some clues or telling me what I'm missing.
Edit: $b\implies c \implies a$ are easy and can be done with the same idea about taking complementaries, I just missed something about the first implication

REPLY [2 votes]: I agree with your idea, show: $\{y\in Y; f^{-1}(y)\subseteq U\} = Y-(f(X-U))$.
Suppose $f^{-1}(y)\subseteq U$. Then nothing in $X-U$ is sent to $y$, i.e. $f^{-1}(y)\cap (X-U) = \emptyset$ so $f(f^{-1}(y))=y\not\in f(X- U)$.
Conversely,
If $y\not \in f(X-U)$ then, again, nothing in $X-U$ is sent to $y$, so $f^{-1}(y)\subseteq U$ (possibly empty!).