Closed map

A closed map sends closed subsets of its domain to closed subsets of its codomain. A bijection is a homeomorphism exactly when it is continuous and closed.

A perfect map is a continuous closed map whose fibers are compact. Surjectivity is sometimes an additional convention; here it is not imposed, and empty fibers are compact. For maps between Tychonoff spaces, the perfect-map characterization by Stone-Čech remainders identifies this property using the unique extension to their compactifications.

For a continuous map $f:X\to Y$ between Tychonoff spaces, extend it to $F:\beta X\to \beta Y$ by the Stone-Čech extension property. If $f$ is perfect, compactness of a fiber and closedness of $f$ show that no remainder point can map to a point of $Y$: separate such a point from the fiber by a neighborhood in $\beta X$, and use closedness to obtain a target neighborhood whose preimage stays inside it, contradicting density. Conversely, if $F^{-1}(Y)=X$, each fiber is compact, and for closed $A\subseteq X$ the equality $f(A)=F({\overline{A}}^{\beta X})\cap Y$ proves closedness. Equivalently $F(\beta X\setminus X)\subseteq \beta Y\setminus Y$.

For metric spaces $X,Y$ with $X$ compact, the projection with a compact factor is closed: if $F\subset X\times Y$ is a closed set and $y_n\in {\pi }_Y(F)$ converges to $y$, choose $x_n$ with $(x_n,y_n)\in F$. Sequential compactness supplies $x_{n_k}\to x$, so $(x_{n_k},y_{n_k})\to (x,y)\in F$, proving $y\in {\pi }_Y(F)$. The projection map is also continuous for the sum metric. Without compactness, the closed set $\{(x,y)\in {\mathbb{R}}^2:xy=1\}$ projects to the nonclosed set $\mathbb{R}\setminus \{0\}$.

Let $f:X\to Y$ be a continuous map that is a closed map and whose fibres are compact. Then $f^{-1}(K)$ is compact for every compact set $K\subseteq Y$, without a Hausdorff hypothesis. Given an open cover, cover each fibre by a finite union $V_y$ of its members. The set $W_y=Y\setminus f(X\setminus V_y)$ is open, contains $y$, and has $f^{-1}(W_y)\subseteq V_y$. A finite subcover of $K$ by the $W_y$ uses only finitely many original cover members and covers $f^{-1}(K)$. Empty fibres cause no exception.