composition of continuous mappings is continuous

Theorem 1.

The composition of two continuous mappings (when defined) is continuous.

Proof.

Let $X, Y, Z$ be topological space, and let $f, g$ be mappings

	$\displaystyle f\colon X$	$\displaystyle\to$	$\displaystyle Y,$
	$\displaystyle g\colon Y$	$\displaystyle\to$	$\displaystyle Z.$

We wish to prove that $g\circ f$ is continuous. Suppose $B$ is an open set in $Z$ . Since $g$ is continuous, $g^{-1}(B)$ is an open set in $Y$ , and since $f$ is continuous, $f^{-1}(g^{-1}(B))$ is an open set in $X$ . Since $f^{-1}(g^{-1}(B))=(g\circ f)^{-1}(B)$ , it follows that $(g\circ f)^{-1}(B)$ is open and the composition if continuous. ∎

Title	composition of continuous mappings is continuous
Canonical name	CompositionOfContinuousMappingsIsContinuous
Date of creation	2013-03-22 15:16:52
Last modified on	2013-03-22 15:16:52
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	8
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 26A15
Classification	msc 54C05