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I was working on this with a group and the group came up with the following for a proof:

Suppose $c \in f^{-1}(f(C))$. Then by definition of $f^{-1}$, $f(c) \in f(C)$. Hence $c \in C$ and $f^{-1}(f(C)) \subseteq C$.

This doesn't seem correct as it doesn't assume the hypothesis. i.e. there is no assumption or use of $f : A \rightarrow B$ and $C \subseteq A$.

Is the proof the group came up with correct and if not any help on the proof would be nice.

Idle Fool
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    The statement is not correct in general, especially if $f$ is not one-to-one. – Solumilkyu Oct 22 '16 at 14:24
  • @Asaf Karagila I looked through the possible questions that came up to ensure that this was not a duplicate. :( I should have thought to just type in the conclusion and see what popped up. – Idle Fool Oct 22 '16 at 15:14
  • It's fine. http://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages is probably the most useful link here. – Asaf Karagila Oct 22 '16 at 15:58

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The result is false. Consider the example $A=B=\Bbb Z$, $f(n)=|n|$, and $C=\Bbb N$ (where my $\Bbb N$ includes $0$). Clearly $f[\Bbb N]=\Bbb N$, so $f^{-1}\big[f[\Bbb N]\big]=f^{-1}[\Bbb N]=\Bbb Z$, and $\Bbb Z\nsubseteq\Bbb N$. In fact the correct inclusion goes the other way: $f^{-1}\big[f[C]\big]\supseteq C$.

Your argument goes astray when you conclude from $f(c)\in f[C]$ that $c\in C$. In my example we have $f(-1)=|-1|=1\in\Bbb N$, but $-1\notin\Bbb N$. If you knew that $f$ was injective (one-to-one), you could legitimately conclude that $c\in C$, but you don’t know that.

Brian M. Scott
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