# Prove that $f(f^{-1}(V))=f^{-1}(f(V))$

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.

• @Dbchatto67 That's not correct, partly because $V$ is not a subset of $X$ and therefore $f(V)$ is undefined. – David Apr 15 at 5:12

## 2 Answers

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 $$(2)$$.

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.