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This question already has an answer here:

Once again I am stuck on a question from Lay's Introduction to Analysis with Proof:

Suppose that $f : A \rightarrow B$ and let C $\subseteq$ A and $D \subseteq B$. Show that if $f$ is injective, then $f^{-1}(f(C))=C$

I just need to show that $f^{-1}(f(C) \subseteq C$ and $C \subseteq f^{-1}(f(C))$.

I have started with this but am unsure of where to go next;

Let $x \in f^{-1}(f(C))$. Then, $x \in \{f^{-1}(a) \mid a \in f(C)\}$.

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marked as duplicate by Claude Leibovici, William, drhab, Davide Giraudo real-analysis Sep 15 '14 at 8:13

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    $\begingroup$ Note that $x\in f^{-1}(A)$ just means that $f(x)\in A$. More details in my answer below. $\endgroup$ – Kim Jong Un Sep 15 '14 at 5:44
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Let $f$ be a one-to-one function. If $x\in C$, then $f^{-1}(f(x))=\{x\}\Rightarrow f^{-1}({f(C)})=C$.

To prove the last implication, this about this. If $g(x)=x$ what is $g(C)$? $C$!

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  • $\begingroup$ You have assumed the proof was true in your proof it seems ... $\endgroup$ – MathMajor Sep 15 '14 at 5:31
  • $\begingroup$ To use this proof don't we first need to prove that $f^{-1}(f(x))=x$ for injective $f$? $\endgroup$ – MathMajor Sep 15 '14 at 5:34
  • $\begingroup$ I think that is clear, since $f(x)$ maps to a unique element $f^{-1}(f(x))=x$. $\endgroup$ – d80d2729a352b1366139fc119d3345 Sep 15 '14 at 5:35
  • $\begingroup$ Yes but $f^{-1}(x)$ only exists if $f$ is one-to-one does it not? $\endgroup$ – MathMajor Sep 15 '14 at 5:36
  • $\begingroup$ For $f^{-1}(x)$ to properly exist over the codomain, we need additional conditions. Here we are talking about $f^{-1}(f(x))$. $\endgroup$ – d80d2729a352b1366139fc119d3345 Sep 15 '14 at 5:39
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If $x\in f^{-1}(f(C))$ then $f(x)\in f(C)$. If $x$ is not in $C$, then there is some element $y\in C$ such that $x\neq y$ and $f(x)=f(y)$ but this violates injectiveness, so it must be that $x\in C$. Therefore, you have one direction of inclusion.

The reverse direction is always true regardless of injectiveness: suppose that $x\in C$, then $f(x)\in f(C)$ so that $x\in f^{-1}(f(C))$.

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Let $f^{-1}(f(C))=B$. Then there exists $x_b\in B$ and $x_c\in C$, such that, $f(x_b)=f(x_c)$. But $f$ is injective so, $x_b=x_c$. Hence $B=C$.

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