Remark 2

Let V be a subset of B. One has $\alpha^{-1}_f(\beta_f(V)) =f(V\cap A)$ and

$\beta^{-1}_f(\beta_f(V)) = V\cup \overset{-1}{f}(f(V\cap A))$. If A is closed in B and if the mapping $f$ is closed, the mapping $\beta_f$ is therefore closed. Analogously, if A is open in B and if the mapping $f$ is open, the mapping $\beta_f$ is open.

In these two cases, the mapping $\beta_f$ then induces a homeomorphism of B-A onto its image.