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George F Simmons, Topology and Modern Analysis pg.79 Problem 4

Let $X$ and $Y$ be metric spaces. Show that an into mapping $f:X \rightarrow Y$ is continuous $\iff$ $f^{-1}\left(G\right)$ is closed in $X$ whenever $G$ is closed in $Y$.

I can prove the problem for open sets, and I have been trying hard for closed. However, seems like I am stuck somewhere missing something obvious. Please don't answer directly, just give a small hint if possible.

EDIT: I am using the definition that $f^{-1}\left(G\right)$ exists only when $f$ is onto and if it is not then $f^{−1}\left(G\right)$ is a loose term for $f^{-1}\left(H\right)$ where $H$ is the range of $f$ in $G$.

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    I was not taught that way, not using that or equivalent definition, I was taught as I wrote in that comment. While, I did learn the new definition, once in$a$while the old habit unconsciously takes over. I hope, I get used to the new one fast enough though :-)2012-07-25

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So you can show that $f$ is continuous iff preimages of open sets are open. Now you go from there. Closed sets are complements of open sets. What is $f^{-1}[Y\setminus G]$?

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    Okay, then it is trivial. You just take the complements and get done. Somehow, now, the question is something else, while the actual difficulty was something else. I guess, I would edit the question and someone can edit his answer and then, I would accept it so that it represents what actually happened?2012-07-24
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The following steps lead to a solution:

(1) $G$ is closed in $Y$ iff $Y - G$ is open in $Y$.

(2) For any $A \subseteq Y$, we have $f^{-1}(Y - A) = X - (f^{-1}(A)).$

(3) Conclude that $f$ is continuous on $X$ $\iff$ $f^{-1}(G)$ is closed in $X$ whenever $G$ is closed in $Y$.

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    @JayeshBadwaik You can define the preimage regardless of what kind of map $f$ is (injective, surjective or neither). (2) that I stated above always holds true.2019-05-25