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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

Community
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Nelson
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1 Answers1

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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!).

David P
  • 12,320
  • Yeah, as I thought it was so easy that I'm ashamed of asking it. I forgot to mention about the others two implications but they are really similar and even more easy, I don't know if I should edit my question giving them. Anyhow everything is done now, thank you a lot! – Nelson Jan 17 '14 at 14:52