n: IT\]) and (\[Eqn: FT’\]) are in terms of inverse images (the first of which constitutes a direct, and the second an inverse, problem) the image $f(U)=\textrm{comp}(V)$ for $V\in\mathcal{V}$ is of interest as it indicates the relationship of the openness of $f$ with its continuity. This, and other related concepts are examined below, where the range space $f(X)$ is always taken to be a subspace of $Y$. Openness of a function *$f\!:(X,\mathcal{U})\rightarrow(Y,\mathcal{V})$* is the “inverse” of continuity, when images of open sets of $X$ are required to be open in $Y$; such a function is said to be *open.* Following are two of the important properties of open functions.

\(1) *If $f\!:(X,\mathcal{U})\rightarrow(Y,f(\mathcal{U}))$ is an open function, then so is* $f_{<}\!:(X,\mathcal{U})\rightarrow(f(X),\textrm{IT}\{ i;f(\mathcal{U})\})$*. The converse is true if $f(X)$ is an open set of $Y$; thus openness of* $f_{<}\!:(X,\mathcal{U})\rightarrow(f(X),f_{<}(\mathcal{U}))$ *implies tha*t *of $f\!:(X,\mathcal{U})\rightarrow(Y,\mathcal{V})$ whenever $f(X)$ is open in $Y$ such that $f_{<}(U)\in\mathcal{V}$ for $U\in\mathcal{U}$.* The truth of this last assertion follows easily from the fact that if $f_{<}(U)$ is an open set of $f(X)\subset Y$, then necessarily $f_{<}(U)=V\bigcap f(X)$ for some $V\in\mathcal{V}$, and the intersection of two open sets of $Y$ is again an open set of $Y$.

\(2) *If $f\!:(X,\mathcal{U})\rightarrow(Y,\mathcal{V})$ and $g\!:(Y,\mathcal{V})\rightarrow(Z,\mathcal{W})$ are open functions then $g\circ f\!:(X,\mathcal{U})\rightarrow(Z,\mathcal{W})$* *is also open.* It follows that the condition in (1) on $f(X)$ can be replaced by the requirement that the inclusion $i\!:(f(X),\textrm{IT}\{ i;\mathcal{V}\})\rightarrow(Y,\mathcal{V})$ be an open map. This interchange of $f(X)$ with its inclusion $i\!:f(X)\rightarrow Y$ into $Y$ is a basic result that finds application in many situations.

Collected below are some useful properties of the initial and final topologies that we need in this work.
