3

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 by $(g\circ f)(w)=g(f(w))$ 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: $\forall B \in \mathcal B(f(x)) \exists A \in \mathcal B(x) : f(A) \subseteq B$
Martin Sleziak
  • 53,687
Jhwana
  • 535

2 Answers2

4

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.

Paul
  • 20,553
2

The part (a) is not answered here - so let me include it, for the sake of completeness.$\newcommand{\inv}[1]{{#1}^{-1}}$

We want to show that $\inv{(g\circ f)}(Z)=\inv f(\inv g(Z))$. For any $w\in W$ we have:

  • $w\in \inv{(g\circ f)}(Z)$ $\Leftrightarrow$ $(g\circ f)(w)\in Z$ $\Leftrightarrow$ $g(f(w))\in Z$
  • $w\in \inv f(\inv g(Z))$ $\Leftrightarrow$ $f(w) \in \inv g(Z)$ $\Leftrightarrow$ $g(f(w))\in Z$

So we see that $$w\in \inv{(g\circ f)}(Z) \Leftrightarrow w\in \inv f(\inv g(Z)),$$ which means that $\inv{(g\circ f)}(Z)=\inv f(\inv g(Z))$.

Proof of this can be found also in an answer to this question: Show that $(g \circ f)^{-1}(C) = g^{-1}(f^{-1}(C)).$ (Although the claim formulated in the title of that question is incorrect - notice the wrong order of $\inv g$ and $\inv f$.)

Martin Sleziak
  • 53,687