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. ) 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. ).
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
A local homeomorphism is an open map.
Let $f \colon X \to Y$ be a local homeomorphism and $U \subset X$ an open subset. We need to see that the image $f(U) \subset Y$ is an open subset of $Y$. For this we may equivalently show that each $y \in f(U)$ has an open neighbourhood inside $f(U)$.
But since any function is surjective onto its image, there exists $x \in U$ with $f(x) = y$. By local homeomorphy of $f$, this $x \in X$ has an open neighbourhood $U_x \subset X$ with $f_{|U_x} \colon U_x \to f(U_x)$ a homeomorphism. Since $U \cap U_x \subset U_x$ is an open neighbourhood of $x$ in $U_x$, the homeomorphy of $f_{|U_x}$ implies that $f(U \cap U_x) \subset f(U)$ is an open neighbourhood of $f(x) = y$.
(preimages of open maps preserve topological interiors)
For $f \,\colon\, X \longrightarrow Y$ an open map and $U \subset X$ a subset, the preimage under $f$ of the interior of $U$ is the interior of the preimage of all of $U$:
In one direction, the inclusion
follows since the left hand side is an open subset of $f^{-1}(U) \subset X$ by continuity of $f$, while the right hand side is the largest such subset, by definition.
In the other direction, the inclusion
is equivalent (see there) to the inclusion
This follows since now the left hand side is an open subset of $f(U)$ by open-ness of $f$, while the right hand side is again the largest open subset of $U$, by definition.
And dually:
(preimages of open maps preserve topological closure)
If $f \,\colon\, X \longrightarrow Y$ is an open map, and $U \subset X$ a subset with topological closure $\overline{U}$, then the preimage $f^{-1}\big( \overline{U}\big)$ of the topological closure is the topological closure of the preimage of $U$:
Noticing that
the topological closure is equivalently the complement of the topological interior of the complement (see there):
preimages evidently preserve complements
it is sufficient to observe that forming preimages of open maps preserves interiors. This is the statement of Prop. .
Last revised on June 30, 2022 at 06:39:42. See the history of this page for a list of all contributions to it.