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Let $f: X\to Y$ be bijective, and $f^{-1}: Y\to X$ be it's inverse. If $V\subseteq Y$, show that the forward image of $V$ under $f^{-1}$ is the same set as the inverse image of $V$ under $f$.

I have interpreted this as: show that $f(f^{-1}(V))=f^{-1}(f(V))$

I really do not know what to do from here.

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    $\begingroup$ @Dbchatto67 That's not correct, partly because $V$ is not a subset of $X$ and therefore $f(V)$ is undefined. $\endgroup$ – David Apr 15 at 5:12
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No, what they want you to show is that the image of $f^{-1} $: $$\tag1 (f^{-1})(V)=\{f^{-1}(v):\ v\in V\} $$ Is equal to the preimage of $V $ under $f $: $$\tag2 f^{-1}(V)=\{x\in X:\ f (x)\in V\}. $$ The notation is unfortunate in this case, but it is normally used for $(2) $.

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Actually, what you have to prove is that $$f^{-1}(V)=f^{-1}(V)\ .$$ To see why this is not obvious, you have to carefully unpack the meanings of all terms, noting in particular that both $f$ and $f^{-1}$ actually have multiple meanings.

First, the inverse image of $V$ under $f$ is by definition $$A=\{\,x\in X\ \mid\ f(x)\in V\,\}\ .$$

Second, what is the forward image of $V$ under $f^{-1}$? We note that since $f$ is a bijection, $f^{-1}$ is a function from $Y$ to $X$ defined by $$f^{-1}(y)=x\quad\hbox{if and only if}\quad f(x)=y\ .$$ The forward image of $V$ under this function is by definition $$B=\{\,f^{-1}(v)\ \mid\ v\in V\,\}\ .$$

So, you have to prove that $A=B$, and now I hope you understand why putting it as "$f^{-1}(V)=f^{-1}(V)$" is so confusing!!

Suppose that $x\in A$. Then $$\eqalign{f(x)=v\quad \hbox{for some $v\in V$}\quad &\Rightarrow\quad x=f^{-1}(v)\quad \hbox{for some $v\in V$}\cr &\Rightarrow\quad x\in B\ ,\cr}$$ so $A\subseteq B$. You also need to show $B\subseteq A$, now that you understand the problem better please try this for yourself.

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