# 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 CompositionOfContinuousMappingsIsContinuous 2013-03-22 15:16:52 2013-03-22 15:16:52 mathcam (2727) mathcam (2727) 8 mathcam (2727) Theorem msc 26A15 msc 54C05