# Prove that if $Z\subseteq Y$, then $(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z)).$

Let $$W ,X$$ and $$Y$$ be three sets and let $$f :W \to X$$ and $$g: X \to Y$$ be two functions. Consider the composition $$g \circ f: W \to Y$$ which, as usual , is defined bt $$(g\circ f)(w)=g(f(w))$$ for $$w \in W$$.

$$(a)$$ Prove that f $$Z\subseteq Y$$, then $$(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z)).$$

$$(b)$$ Deduce that if $$(W,c) ,(X,d)$$ and $$(Y,e)$$ are metric spaces and the functions $$f$$ and $$g$$ are both continuous ,then the function $$g \circ f$$ is continuous.

Definitions:

• Let $$(X, d)$$ and $$(Y, e)$$ be metric spaces, and let$$x \in X$$. A function $$f : X \to Y$$is continuous at $$x$$ if: $$\forall B \in \mathcal B(f(x)) \exists A \in \mathcal B(x) : f(A) \subseteq B$$
• Do you know the definitions? If so, what have you tried? – Clayton Apr 13 '13 at 12:32
• These are trivil. I believe you could solve them by yourself. – Paul Apr 13 '13 at 12:43
• Is math.stackexchange.com/a/2426/60329 correct for (a)? – Jhwana Apr 13 '13 at 12:45
• @Fayz: I have given an answer to the question b). – Paul Apr 13 '13 at 13:18
• @HenrySwanson: It means the set of all open balls with centerd $x$. – Jhwana Apr 13 '13 at 13:20

Note that $f: X \rightarrow Y$ is continuous iff for for any open set $U \subseteq Y$, $f^{-1}(U)$ is open in $X$.
To prove $g \circ f$ is continuous, for any open set $U \subset Y$, we only need to prove $(g \circ f)^{-1}(U)$ is open in $W$.
To see this, as $(g \circ f)^{-1}(U)=f^{-1}(g^{-1}(U))$, and $g$ is continuous, we see $g^{-1}(U)$ is open; as $f$ is continuous, then $f^{-1}(g^{-1}(U))$ is also open.