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Basic question regarding function.

Let $f: X \to Y$, then for what $f$ does $f(f^{-1}(f(A))) = f(A)$? hold?

Obviously this relationship holds when $f$ is a bijection.

This does not hold when $f$ is pure surjection because the inverse does not exist.

Does this also hold when $f$ is an pure injection? I think so.

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    $\begingroup$ Are you sure $f^{-1}(S)$ means the inverse function of $f$ applied to $S$, rather than the preimage of $S$? $\endgroup$
    – pjs36
    Jun 9, 2016 at 20:35
  • $\begingroup$ @pjs36 In all actuality I am totally not sure.. $\endgroup$
    – Olórin
    Jun 9, 2016 at 22:33
  • $\begingroup$ Possible duplicate of Prove that $f[f^{-1} f[X]]] = f[X]$ $\endgroup$
    – Watson
    Jan 26, 2017 at 12:29

3 Answers 3

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Actually it always holds.

But $f^{-1}$ must not be considered as a function $Y\to X$. Instead, $f^{-1}(B)$ is defined for every $B\subset Y$ as $$ f^{-1}(B) = \{x\in X| f(x)\in B\}.$$ With this definition, for any $x\in A$, $x\in f^{-1}(f(A))$, so that the inclusion $f(A)\subset f(f^{-1}(f(A)))$ holds. Moreover, if $x\in f^{-1}(f(A))$, then $f(x)\in f(A)$, and thus the other inclusion holds. Thus we have the equality $f(A)= f(f^{-1}(f(A)))$.

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  • $\begingroup$ @kccu thanks for the correction. $\endgroup$
    – Arnaud D.
    Jun 9, 2016 at 20:45
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When $f$ is an injection function then $f^{-1}(f(A))=A$, as a result $$f(f^{-1}(f(A))) = f(A)$$

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I would assume here that $f^{-1}(S)$, for $S \subseteq Y$, is referring to the preimage of $S$ under $f$. To avoid confusion with the inverse function $f^{-1} : Y \to X$ (should such a thing exist), I'll follow the somewhat common convention to use square brackets when taking preimages, writing $f^{-1}[S]$ for $\{x \in X : f(x) \in S\}$ (this helps prevent mistaking inverse functions for preimages).

$f^{-1}[f(A)]$ is by definition the set $\{x \in X: f(x) \in f(A)\}$. Applying $f$ to this set (that is, to each element in the set) yields $$f(f^{-1}[f(A)]) = \{f(x) : f(x) \in f(A)\},$$

and this is $f(A)$.

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  • $\begingroup$ I tend to prefer using the superscript arrows for image and preimage maps, thus I would write $f^{\rightarrow}(A)$ for the image under $f$ of $A$, and $f^{\leftarrow}(B)$ for the preimage under $f$ of $B$. $\endgroup$ Jun 9, 2016 at 21:01
  • $\begingroup$ Sources that use $f^{-1}[S]$ for preimage use $f[S]$ for image, I would think. $\endgroup$
    – GFauxPas
    Jun 9, 2016 at 21:05
  • $\begingroup$ @JustinBenfield That's not a bad idea, although I can't say I've seen it before. $\endgroup$
    – pjs36
    Jun 9, 2016 at 21:14
  • $\begingroup$ @GFauxPas Yes, I think that's likely true. I kind of liked the "legibility" aspect of alternating symbols here, so kept the round parentheses for images. $\endgroup$
    – pjs36
    Jun 9, 2016 at 21:15
  • $\begingroup$ @pjs36 I was originally introduced to that notation by my MTH-311 professor, but I doubt he originated the idea. What's nice about it is that it's entirely unambiguous whereas writing $f^{-1}$ for anything other than the inverse of a function is equivocating on the notation. The arrow notation is apparently well enough known that it's on the wikipedia page for image: en.wikipedia.org/wiki/Image_(mathematics) $\endgroup$ Jun 9, 2016 at 22:21

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