Remark 1

The mapping $\alpha_f$ is injective. For every subset U

of X, one has $\alpha^{-1}_f(\alpha_f(U)) = U$ and $\beta^{-1}_f(\alpha_f(U)) =\overset{-1}{f}(U)$. If A is an open (resp. closed) subset of B, these relations imply that $\alpha_f$ is an open (resp. closed) mapping of X into $X\cup_fB$.

Let U be an open subset of X and let V be an open subset of B such that $\overset{-1}{f}(U) = V\cap A$. The union of the subsets U of $Y_1$ and V of $Y_2$ is a saturated open subset of Y; its image in $X\cup_fB$ is therefore open, and its trace on $\alpha_f(X)$ is $\alpha_f(U)$. It follows that the mapping $\alpha_f$ defines a homeomorphism of X onto its image in $X\cup_fB$.