If $f: A \rightarrow B$ and $g: B \rightarrow C$, is $(g \circ f)^{-1}(C) = f^{-1}(B)$?

Question

Let $f: A \rightarrow B$ and $g: B \rightarrow C$. I have a question that is asking me to come up with and prove some theorems about the images and inverse images of sets under $g \circ f$. This is a theorem and proof that I have so far (sorry, I'm not great at writing proofs):

Theorem: Suppose $f: A \rightarrow B$ and $g: B \rightarrow C$. Then $(g \circ f)^{-1}(C) = f^{-1}(B)$.

$(\subseteq)$ Suppose $x \in (g \circ f)^{-1}(C)$. This implies that $x \in A$. Since $x \in A$ and $f: A \rightarrow B$ is a function, then $f(x) \in B$. Thus, since $x \in A$ and $f(x) \in B$, it follows that $x \in f^{-1}(B)$.

$(\supseteq)$ Suppose $x \in f^{-1}(B)$. This implies that $x \in A$ and $f(x) \in B$. Since $f(x) \in B$ and $g: B \rightarrow C$ is a function, then $g(f(x)) \in C$. Thus, since $x \in A$ and $g(f(x))=(g \circ f)(x) \in C$, it follows that $x \in (g \circ f)^{-1}(C)$.

Is this a valid theorem and proof, or is this not always true? I'm having trouble trying to come up with a counterexample.

Answer 1

Provided that $f,g$ are both $1-1$ and unto, you can also work this out by first showing that $$(g \circ f)^{-1}(x)=f^{-1}\circ g^{-1}(x).$$ Then it follows that $$(g\circ f)^{-1}(C)=f^{-1}(g^{-1}(C))=f^{-1}(B) $$

Answer 2

Your proof is valid. However, it would be easier to prove that if $h : X \to Y$ is any function then $h^{-1}(Y) = X$. It then follows that $(g \circ f)^{-1}(C) = f^{-1}(B)$ simply because $g \circ f$ and $f$ have the same domain!