Topology and groupoid, 7.3.4, push out of cofibre pair.

Question

In Ronnie Brown's, pg 278, 7.3.4 he stated

Let $(X,A)$ be a closed fibred pair, and let $f:A \rightarrow B$ be a map. Then $(B{_f} \sqcup X , B)$ is a closed fibred pair.

Why do we need the closed condition? I do not see this used anywhere in the proof. Unless I am missing something?

Answer 1

You are right, it is not needed. The pushout of a cofibration is always a cofibration. This follows from the universal property of pushout diagrams. However, Ronnie Brown probably wants to emphasize that you obtain a closed cofibred pair.