Remark 1

Let A be a subset of $X\times \mathbf{I}$ and let $A_1$ be the set of points $x\in X$ such that $(x,1)\in A$. We have $\alpha^{-1}_f(\alpha_f(A)) = A\cup \overset{-1}{f}(f(A_1))\times \{1\}$ and $\beta^{-1}_f(\alpha_f(A)) =f(A_1)$. It follows that the mapping $\alpha_f$ is closed (resp. open) if $f$ is.