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While reading about inverse functions I found a statement showing "$f\{f^{-1}(x)\}=x$". I know it is correct but I tried to prove it by taking an simple example.

Suppose there is a function $y=f(x)$ given as, $f:$ $\ce{A=B}$. Here $A$ and $B$ represents domain and range respectively and are sets given as $A=\{1\}$ & $B=\{2\}$, then $f(x)$ can be also written as $f(x)=\{(1,2)\}$. So for $x=1$ (domain), $y=2$ (range). Where $1$ belongs to $A$ and $2$ belongs to $B$.

Also $f^{-1}(x)=\{(2,1)\}$. Now $f\{f^{-1}(x)\}=\{(2,2)\}$, i.e. identity function of range $y$, i.e. set $B$. As in identity function range $=$ domain. Therefore $y=x$. Hence $f\{f^{-1}(x)\}=x$ Am I correct in explaining this? As I think my method of explanation is wrong. Please tell me the right one & please point out my mistakes.

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    You don't have to prove $f(f^{-1}(x)) = x$, it's literally part of the definition of $f^{-1}$ (the other part being that $f^{-1}(f(x))$ has to equal $x$ as well). – 2017-02-20
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    But sure, your example looks fine (other than some notational issues). – 2017-02-20

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Let $f:X \to Y$ and $A \subseteq Y$. The preimage of a set is defined as follows:

$$f^{-1}(A)=\{x \in X \mid f(x) \in A\}.$$

This means that the preimage of $\{x\}$ (i. e. $f^{-1}(\{x\})$ are exactly the elements so that $f$ applied them lands you in $\{x\}$, as in $f(f^{-1}(\{x\}))=x.$

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Consider $f:A\longrightarrow B$ such that given $a\in A$, $f(a)=b$ and suppose there exists $f^{-1}:B\longrightarrow A$ that, given $b\in B$, $f^{-1}(b)=a$.

We define the identity function as follows: $$Id_{A}:A\longrightarrow A$$ $$a\mapsto Id_{A}(a)= a$$ $$Id_{B}:B\longrightarrow B$$ $$b\mapsto Id_{B}(b)= b$$

If we define $$f(f^{-1}):B\longrightarrow B$$ $$b\mapsto f(f^{-1}(b))=f(a)=b$$ And $$f^{-1}(f):A\longrightarrow A$$ $$a\mapsto f^{-1}(f(a))=f^{-1}(b)=a$$

Therefore, the composition of inverses acts like $Id_{A}$ and like $Id_{B}$. As they are defined on the same domain, we have that:

$$f(f^{-1})=Id_{B}$$ $$f^{-1}(f)=Id_{A}$$

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    This isn't correct, or at least isn't clearly stated. The phrase "Consider $f:A \to B$ such that given $a\in A$, $f(a)=b$" is meaningless because the variable $b$ has no defined meaning there. Likewise, the phrase "suppose there exists $f^{-1}:B \to A$ that, given $b\in B$, $f^{-1}(b)=a$" is meaningless because the variable $a$ has no defined meaning there. Later on, it kind of feels like you're using $a$ and $b$ to mean "the appropriate element of $A$" and "the appropriate element of $B$", respectively, but it's not clear what those elements are. – 2017-02-20
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    @TannerSwett I believe it is quite clear that $b\in B$ and that $a\inA$, as it follows immediately from the definition of $f$ and $f^{-1}$ respectively. Nevertheless, I appreciate your correction, although it might have been more polite not to begin your comment with "This isn't correct". – 2017-02-20
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    Sorry, you're right, I shouldn't have said that. In any case, by "given $a \in A$, $f(a) = b$", did you mean "given $a \in A$, there exists $b \in B$ such that $f(a) = b$"? I don't understand what your "such that" is saying about $f$. – 2017-02-20
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    @TannerSwett Thats okay, don't worry ;). Yes, I mean there exists. I believe the problem is a did a bad translation here. – 2017-02-20