This page is about the concept in topology. For the more general concept see at open morphism.
topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
(open maps and closed maps)
A continuous function $f \colon (X,\tau_X) \to (Y, \tau_Y)$ between topological spaces is called
an open map if the image under $f$ of an open subset of $X$ is an open subset of $Y$;
a closed map if the image under $f$ of a closed subset of $X$ is a closed subset of $Y$.
(image projections of open/closed maps are themselves open/closed)
If a continuous function $f \colon (X,\tau_X) \to (Y,\tau_Y)$ is an open map or closed map (def. 1) then so is its image projection $X \to f(X) \subset Y$, respectively, for $f(X) \subset Y$ regarded with its subspace topology.
If $f$ is an open map, and $O \subset X$ is an open subset, so that $f(O) \subset Y$ is also open in $Y$, then, since $f(O) = f(O) \cap f(X)$, it is also still open in the subspace topology, hence $X \to f(X)$ is an open map.
If $f$ is a closed map, and $C \subset X$ is a closed subset so that also $f(C) \subset Y$ is a closed subset, then the complement $Y \backslash f(C)$ is open in $Y$ and hence $(Y \backslash f(C)) \cap f(X) = f(X) \backslash f(C)$ is open in the subspace topology, which means that $f(C)$ is closed in the subspace topology.
(projections out of product spaces are open maps)
For $(X_1,\tau_{X_1})$ and $(X_2,\tau_{X_2})$ two topological spaces, then the projection maps
out of their product topological space
are open continuous functions (def. 1).
This is because, by definition, every open subset $O \subset X_1 \times X_2$ in the product space topology is a union of products of open subsets $U_i \in X_1$ and $V_i \in X_2$ in the factor spaces
and because taking the image of a function preserves unions of subsets