Open and closed maps

If S is a subset of a topological space then let \overline{S} and \operatorname{Cl} S (resp. \operatorname{Int} S) denote the closure (resp. interior) of S in that space. Let f : X \to Y be a function between topological spaces. If S is any set then f(S) := \left\{ f(s) ~:~ s \in S \cap \operatorname{domain} f \right\} is called the image of S under f. Competing definitions There are two different competing, but closely related, definitions of "" that are widely used, where both of these definitions can be summarized as: "it is a map that sends open sets to open sets." The following terminology is sometimes used to distinguish between the two definitions. A map f : X \to Y is called a • "''''''" if whenever U is an open subset of the domain X then f(U) is an open subset of f's codomain Y. • "''''''" if whenever U is an open subset of the domain X then f(U) is an open subset of f's image \operatorname{Im} f := f(X), where as usual, this set is endowed with the subspace topology induced on it by f's codomain Y. Every strongly open map is a relatively open map. However, these definitions are not equivalent in general. :Warning: Many authors define "open map" to mean " open map" (for example, The Encyclopedia of Mathematics) while others define "open map" to mean " open map". In general, these definitions are equivalent so it is thus advisable to always check what definition of "open map" an author is using. A surjective map is relatively open if and only if it is strongly open; so for this important special case the definitions are equivalent. More generally, a map f : X \to Y is relatively open if and only if the surjection f : X \to f(X) is a strongly open map. Because X is always an open subset of X, the image f(X) = \operatorname{Im} f of a strongly open map f : X \to Y must be an open subset of its codomain Y. In fact, a relatively open map is a strongly open map if and only if its image is an open subset of its codomain. In summary, :A map is strongly open if and only if it is relatively open and its image is an open subset of its codomain. By using this characterization, it is often straightforward to apply results involving one of these two definitions of "open map" to a situation involving the other definition. The discussion above will also apply to closed maps if each instance of the word "open" is replaced with the word "closed". Open maps A map f : X \to Y is called an ' or a ' if it satisfies any of the following equivalent conditions: Definition: f : X \to Y maps open subsets of its domain to open subsets of its codomain; that is, for any open subset U of X, f(U) is an open subset of Y. f : X \to Y is a relatively open map and its image \operatorname{Im} f := f(X) is an open subset of its codomain Y. For every x \in X and every neighborhood N of x (however small), f(N) is a neighborhood of f(x). We can replace the first or both instances of the word "neighborhood" with "open neighborhood" in this condition and the result will still be an equivalent condition: • For every x \in X and every open neighborhood N of x, f(N) is a neighborhood of f(x). • For every x \in X and every open neighborhood N of x, f(N) is an open neighborhood of f(x). f\left( \operatorname{Int}_X A \right) \subseteq \operatorname{Int}_Y ( f(A) ) for all subsets A of X, where \operatorname{Int} denotes the topological interior of the set. Whenever C is a closed subset of X then the set \left\{ y \in Y ~:~ f^{-1}(y) \subseteq C \right\} is a closed subset of Y. • This is a consequence of the identity f(X \setminus R) = Y \setminus \left\{ y \in Y : f^{-1}(y) \subseteq R \right\}, which holds for all subsets R \subseteq X. If \mathcal{B} is a basis for X then the following can be appended to this list: • f maps basic open sets to open sets in its codomain (that is, for any basic open set B \in \mathcal{B}, f(B) is an open subset of Y). Closed maps A map f : X \to Y is called a '''''' if whenever C is a closed subset of the domain X then f(C) is a closed subset of f's image \operatorname{Im} f := f(X), where as usual, this set is endowed with the subspace topology induced on it by f's codomain Y. A map f : X \to Y is called a ' or a ' if it satisfies any of the following equivalent conditions: Definition: f : X \to Y maps closed subsets of its domain to closed subsets of its codomain; that is, for any closed subset C of X, f(C) is a closed subset of Y. f : X \to Y is a relatively closed map and its image \operatorname{Im} f := f(X) is a closed subset of its codomain Y. \overline{f(A)} \subseteq f\left(\overline{A}\right) for every subset A \subseteq X. \overline{f(C)} \subseteq f(C) for every closed subset C \subseteq X. Whenever U is an open subset of X then the set \left\{y \in Y ~:~ f^{-1}(y) \subseteq U\right\} is an open subset of Y. If x_{\bull} is a net in X and y \in Y is a point such that f\left(x_{\bull}\right) \to y in Y, then x_{\bull} converges in X to the set f^{-1}(y). • The convergence x_{\bull} \to f^{-1}(y) means that every open subset of X that contains f^{-1}(y) will contain x_j for all sufficiently large indices j. A surjective map is strongly closed if and only if it is relatively closed. So for this important special case, the two definitions are equivalent. By definition, the map f : X \to Y is a relatively closed map if and only if the surjection f : X \to \operatorname{Im} f is a strongly closed map. If in the open set definition of "continuous map" (which is the statement: "every preimage of an open set is open"), both instances of the word "open" are replaced with "closed" then the statement of results ("every preimage of a closed set is closed") is to continuity. This does not happen with the definition of "open map" (which is: "every image of an open set is open") since the statement that results ("every image of a closed set is closed") is the definition of "closed map", which is in general equivalent to openness. There exist open maps that are not closed and there also exist closed maps that are not open. This difference between open/closed maps and continuous maps is ultimately due to the fact that for any set S, only f(X \setminus S) \supseteq f(X) \setminus f(S) is guaranteed in general, whereas for preimages, equality f^{-1}(Y \setminus S) = f^{-1}(Y) \setminus f^{-1}(S) always holds. ==Examples==

The function f : \R \to \R defined by f(x) = x^2 is continuous, closed, and relatively open, but not (strongly) open. This is because if U = (a, b) is any open interval in f's domain \R that does contain 0 then f(U) = (\min \{ a^2, b^2 \}, \max \{ a^2, b^2 \}), where this open interval is an open subset of both \R and \operatorname{Im} f := f(\R) = [0, \infty). However, if U = (a, b) is any open interval in \R that contains 0 then f(U) = [0, \max \{ a^2, b^2 \}), which is not an open subset of f's codomain \R but an open subset of \operatorname{Im} f = [0, \infty). Because the set of all open intervals in \R is a basis for the Euclidean topology on \R, this shows that f : \R \to \R is relatively open but not (strongly) open. If Y has the discrete topology (that is, all subsets are open and closed) then every function f : X \to Y is both open and closed (but not necessarily continuous). For example, the floor function from \R to \Z is open and closed, but not continuous. This example shows that the image of a connected space under an open or closed map need not be connected. Whenever we have a product of topological spaces X=\prod X_i, the natural projections p_i : X \to X_i are open (as well as continuous). Since the projections of fiber bundles and covering maps are locally natural projections of products, these are also open maps. Projections need not be closed, however. Consider for instance the projection p_1 : \R^2 \to \R on the first component; then the set A = \{(x, 1/x) : x \neq 0\} is closed in \R^2, but p_1(A) = \R \setminus \{0\} is not closed in \R. However, for a compact space Y, the projection X \times Y \to X is closed. This is essentially the tube lemma. To every point on the unit circle we can associate the angle of the positive x-axis with the ray connecting the point with the origin. This function from the unit circle to the half-open interval [0,2π) is bijective, open, and closed, but not continuous. It shows that the image of a compact space under an open or closed map need not be compact. Also note that if we consider this as a function from the unit circle to the real numbers, then it is neither open nor closed. Specifying the codomain is essential. ==Sufficient conditions==

Every homeomorphism is open, closed, and continuous. In fact, a continuous bijection is a homeomorphism if and only if it is open, or equivalently, if and only if it is closed. The composition of two (strongly) open maps is an open map and the composition of two (strongly) closed maps is a closed map. However, the composition of two relatively open maps need not be relatively open and the composition of two relatively closed maps need not be relatively closed. If f : X \to Y is strongly open (respectively, strongly closed) and g : Y \to Z is relatively open (respectively, relatively closed), then g \circ f : X \to Z is relatively open (respectively, relatively closed). Let f : X \to Y be a map. Given any subset T \subseteq Y, if f is relatively open (respectively, relatively closed, strongly open, strongly closed, continuous, surjective), then the same is true of its restriction f\big\vert_{f^{-1}(T)} ~:~ f^{-1}(T) \to T to the f-saturated subset f^{-1}(T). The categorical sum of two open maps is open, and of two closed maps is closed. The categorical product of two open maps is also open. However, the categorical product of two closed maps need not be closed. A bijective map is open if and only if it is closed. The inverse of a continuous bijection is an open and closed bijection (and vice versa). An open surjection is not necessarily closed, and a closed surjection is not necessarily open. All local homeomorphisms, including all coordinate charts on manifolds and all covering maps, are open maps. A variant of the closed map lemma states that if a continuous function between locally compact Hausdorff spaces is proper, then it is also closed. In complex analysis, the identically named open mapping theorem states that every non-constant holomorphic function defined on a connected open subset of the complex plane is an open map. The invariance of domain theorem states that a continuous and locally injective function between two n-dimensional topological manifolds must be open. In functional analysis, the open mapping theorem states that every continuous linear surjection between Banach spaces is an open map. This theorem has been generalized to topological vector spaces beyond just Banach spaces. A surjective map f : X \to Y is called an ' if, for every y \in Y, there exists some x \in f^{-1}(y) such that x is a ' for f which by definition means that, for every open neighborhood U of x, f(U) is a neighborhood of f(x) in Y (note that the neighborhood f(U) is not required to be an neighborhood). Every open surjection is an almost open map, but the converse is false. If a surjection f : (X, \tau) \to (Y, \sigma) is an almost open map, then it is an open map if it satisfies the following condition (a condition that does depend in any way on Y's topology \sigma): whenever m and n \in X belong to the same fiber of f (that is, f(m) = f(n)), then, for every neighborhood U \in \tau of m, there exists some neighborhood V \in \tau of n such that F(V) \subseteq F(U). If the map is continuous, then the above condition is also necessary for the map to be open. That is, if f : X \to Y is a continuous surjection, then it is open if and only if it is almost open and it satisfies the above condition. ==Properties==

Open or closed maps that are continuous If f : X \to Y is a continuous map that is also open closed, then: • if f is a surjection, then it is a quotient map and even a hereditarily quotient map, • A surjective map f : X \to Y is called if for every subset T \subseteq Y, the restriction f\big\vert_{f^{-1}(T)} ~:~ f^{-1}(T) \to T is a quotient map. • if f is an injection, then it is a topological embedding. • if f is a bijection, then it is a homeomorphism. In the first two cases, being open or closed is merely a sufficient condition for the conclusion that follows. In the third case, it is necessary as well. Open continuous maps If f : X \to Y is a continuous (strongly) open map, A \subseteq X, and S \subseteq Y, then: f^{-1}\left(\operatorname{Bd}_Y S\right) = \operatorname{Bd}_X \left(f^{-1}(S)\right) where \operatorname{Bd} denotes the boundary of a set. f^{-1}\left(\overline{S}\right) = \overline{f^{-1}(S)} where \overline{S} denote the closure of a set. If \overline{A} = \overline{\operatorname{Int}_X A}, where \operatorname{Int} denotes the interior of a set, then \overline{\operatorname{Int}_Y f(A)} = \overline{f(A)} = \overline{f\left(\operatorname{Int}_X A\right)} = \overline{f \left(\overline{\operatorname{Int}_X A}\right)} where this set \overline{f(A)} is also necessarily a regular closed set (in Y). ==See also==