Proof Verification: $(g\circ f)^{-1}(H)$ = $f^{-1}g^{-1}(H)$ 
Theorem: $ f:A \to B$ and $g: B \to C$ are functions and $H \subset
 C$. Then prove that : 
$(g\circ f)^{-1}(H) = f^{-1}(g^{-1}(H))$

Proof: 
For some $x \in A$, $(g\circ f)(x) = h$ for some $h \in H$
Then: $f^{-1}g^{-1}(h) = f^{-1}g^{-1}((g\circ f)(x))$
$\iff$  $f^{-1}g^{-1}(g(f(x))$  
Since $g^{-1}g(f(x))=f(x)$ by property of inverses
$\iff$ $f^{-1}(f(x))$
$\iff$ $x$
This implies $(g\circ f)^{-1}(H)$ = $f^{-1}g^{-1}(H)$
Is this proof correct? Also, is my usage of $\iff$ correct or would $\implies$ be more appropriate? If yes, then why? 
Thank you. 
 A: Something is going wrong here. 
Suppose we are given a function $f\colon A \to B$ and $C \subset B$. Then the pre-image of $C$, $f^{-1}[C]$, is defined as $\{x \in A \mid f(x) \in C\}$. But note:

This definition makes perfect sense even if the function $f$ has no inverse. In other words, there may well be no inverse function $f^{-1}\colon B \to A$. Indeed, usually there won't be.

You might think (rightly!) that this is an unfortunate notational glitch, this double use of $f^{-1}$, in symbolism for inverses and for pre-images.  Sadly, it is one you have to learn to live with!
So what you are being asked to prove is that $(g \circ f)^{-1}[H] = f^{-1}[g^{-1}[H]]$; and you must do this without supposing there are functions $f^{-1}$, $g^{-1}$ or $(g \circ f)^{-1}$. But it is easy. Just use the definition of pre-image as given. You need to show that if $x$ is one set it is in the other, in both directions. Laura's answer indicates how. 

Helpful hint. Do use the (pretty standard) square-bracket notation for pre-images: i.e. use $f^{-1}[C]$ rather than $f^{-1}(C)$. Then you won't be so tempted to confuse '$f^{-1}$' here with its inverse function usage. 

A: I think that you are mixing the definition of $g^{-1}$. It is not the inverse of the function $g$, but you are working with preimages. You need to prove both inclusions.
For the first one, suppose that $x\in (g\circ f)^{-1}(H)$. Then, by definition, $g(f(x))\in H$. That is $f(x)\in g^{-1}(H)$, so $x\in f^{-1}(g^{-1}(H))$.
The other inclusion is proved similarly. Tell me if you do not understand something.
