nLab open map

This page is about the concept in topology. For the more general concept see at open morphism.

Definition

Definition

(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 $Y$.

Examples

Example

(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.

Proof

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.

Example

(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

$$pr_i \;\colon\; (X_1 \times X_2, \tau_{X_1 \times X_2}) \longrightarrow (X_i, \tau_{X_i})$$

out of their product topological space

$$\array{ X_1 \times X_2 &\overset{pr_1}{\longrightarrow}& X_1 \\ (x_1, x_2) &\overset{\phantom{AAA}}{\mapsto}& x_1 }$$

$$\array{ X_1 \times X_2 &\overset{pr_2}{\longrightarrow}& X_2 \\ (x_1, x_2) &\overset{\phantom{AAA}}{\mapsto}& x_2 }$$

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

$$O = \underset{i \in I}{\cup} \left( U_i \times V_i \right)$$

and because taking the image of a function preserves unions of subsets

$$\begin{aligned} pr_1\left( \underset{i \in I}{\cup} \left( U_i \times V_i \right) \right) & = \underset{i \in I}{\cup} pr_1 \left( U_i \times V_i \right) \\ & = \underset{i \in I}{\cup} U_i \end{aligned} \,.$$

Proposition

A local homeomorphism is an open map.

Proof

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 $f(x) = y$.

Properties

Proposition

(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 $U$:

$$\big( f^{-1}(U) \big)^\circ \;=\; f^{-1}\big( U^\circ\big) \,.$$

Proof

In one direction, the inclusion

$$f^{-1}\big( U^\circ\big) \;\subset\; \big( f^{-1}(U) \big)^\circ$$

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

$$\big( f^{-1}(U) \big)^\circ \;\subset\; f^{-1}\big( U^\circ\big)$$

is equivalent (see there) to the inclusion

$$f\Big( \big( f^{-1}(U) \big)^\circ \Big) \;\subset\; U^\circ \,.$$

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:

Proposition

(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 $U$:

$$\overline{f^{-1}(U)} \;=\; f^{-1}\big( \overline{U} \big) \,.$$

Proof

Noticing that

1. the topological closure is equivalently the complement of the topological interior of the complement (see there):

   $$\overline{U} \,=\, X \setminus (X \setminus U)^\circ$$

2. preimages evidently preserve complements

it is sufficient to observe that forming preimages of open maps preserves interiors. This is the statement of Prop. .

Related concepts

• open morphism

• embedding of topological spaces

• closed map, proper map