# 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

## 1 Answer

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.