Almost open map

In functional analysis and related areas of mathematics, an almost open map between topological spaces is a map that satisfies a condition similar to, but weaker than, the condition of being an open map.
As described below, for certain broad categories of topological vector spaces, surjective linear operators are necessarily almost open.

Definitions

Given a surjective map $f : X \to Y,$ a point $x \in X$ is called a for $f$ and $f$ is said to be (or ) if 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).

A surjective map is called an if it is open at every point of its domain, while it is called an each of its fibers has some point of openness.
Explicitly, a surjective map $f : X \to Y$ is said to be if for every $y \in Y,$ there exists some $x \in f^(y)$ such that $f$ is open at $x.$
Every almost open surjection is necessarily a (introduced by Alexander Arhangelskii in 1963), which by definition means that for every $y \in Y$ and every neighborhood $U$ of $f^(y)$ (that is, $f^(y) \subseteq \operatorname_X U$), $f(U)$ is necessarily a neighborhood of $y.$

Almost open linear map

A linear map $T : X \to Y$ between two topological vector spaces (TVSs) is called a or an if for any neighborhood $U$ of $0$ in $X,$ the closure of $T(U)$ in $Y$ is a neighborhood of the origin.
Importantly, some authors use a different definition of "almost open map" in which they instead require that the linear map $T$ satisfy: for any neighborhood $U$ of $0$ in $X,$ the closure of $T(U)$ in $T(X)$ (rather than in $Y$) is a neighborhood of the origin;
this article will not use this definition.

If a linear map $T : X \to Y$ is almost open then because $T(X)$ is a vector subspace of $Y$ that contains a neighborhood of the origin in $Y,$ the map $T : X \to Y$ is necessarily surjective.
For this reason many authors require surjectivity as part of the definition of "almost open".

If $T : X \to Y$ is a bijective linear operator, then $T$ is almost open if and only if $T^$ is almost continuous.

Relationship to open maps

Every surjective open map is an almost open map but in general, the converse is not necessarily true.
If a surjection $f : (X, \tau) \to (Y, \sigma)$ is an almost open map then it will be an open map if it satisfies the following condition (a condition that does depend in any way on $Y$'s topology $\sigma$):
:whenever $m, n \in X$ 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 an open map if and only if it is almost open and it satisfies the above condition.

Open mapping theorems

:Theorem: If $T : X \to Y$ is a surjective linear operator from a locally convex space $X$ onto a barrelled space $Y$ then $T$ is almost open.

:Theorem: If $T : X \to Y$ is a surjective linear operator from a TVS $X$ onto a Baire space $Y$ then $T$ is almost open.

The two theorems above do require the surjective linear map to satisfy topological conditions.

:Theorem: If $X$ is a complete pseudometrizable TVS, $Y$ is a Hausdorff TVS, and $T : X \to Y$ is a closed and almost open linear surjection, then $T$ is an open map.

:Theorem: Suppose $T : X \to Y$ is a continuous linear operator from a complete pseudometrizable TVS $X$ into a Hausdorff TVS $Y.$ If the image of $T$ is non- meager in $Y$ then $T : X \to Y$ is a surjective open map and $Y$ is a complete metrizable space.

See also

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References

Bibliography

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{{TopologicalVectorSpaces

Topological vector spaces