Proposition 2

Let $X, Y$ be two topological spaces, $f$ a mapping of $X$ into $Y$. For each subset $T$ of $Y$ let $f_T$ denote the mapping of $f^{-1}(T)$ into $T$ which agrees with $f$ on $f^{-1}(T)$.

a) If $f$ is open (resp. closed), $f_T$ is open (resp. closed).

b) Let $(T(i))_{i \in I}$ be a family of subsets of $Y$ whose interiors cover $Y$, or which is a locally finite closed covering of $Y$; if all the $f_{T(i)}$ are open (resp. closed), then $f$ is open (resp. closed).

a) If $A$ is an open (resp. closed) subset of $f^{-1}(T)$, then there is an open (resp. closed) subset $B$ of $X$ such that $A = B \cap f^{-1}(T)$, and therefore f_T(A) = f(B) \cap T; by hypothesis, $f(B)$ is open (resp. closed), so that $f_T(A)$ is open (resp. closed) in $T$.

b) Let $B$ be an open (resp. closed) subset of $X$, and let $B_i$ denote $B \cap f^{-1}(T(i))$; then $f(B) \cap T(i) = f_{T(i)}(B_i)$. Since $f_{T(i)}(B_i)$ is open (resp. closed) in $T(i)$ by hypothesis, it follows that $f(B)$ is open (resp. closed) in $Y$, by Proposition 3 of § 3, no. 1.