nLab open map


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


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topological homotopy theory




(open maps and closed maps)

A continuous function f:(X,τ X)(Y,τ Y)f \colon (X,\tau_X) \to (Y, \tau_Y) between topological spaces is called



(image projections of open/closed maps are themselves open/closed)

If a continuous function f:(X,τ X)(Y,τ Y)f \colon (X,\tau_X) \to (Y,\tau_Y) is an open map or closed map (def. ) then so is its image projection Xf(X)YX \to f(X) \subset Y, respectively, for f(X)Yf(X) \subset Y regarded with its subspace topology.


If ff is an open map, and OXO \subset X is an open subset, so that f(O)Yf(O) \subset Y is also open in YY, then, since f(O)=f(O)f(X)f(O) = f(O) \cap f(X), it is also still open in the subspace topology, hence Xf(X)X \to f(X) is an open map.

If ff is a closed map, and CXC \subset X is a closed subset so that also f(C)Yf(C) \subset Y is a closed subset, then the complement Y\f(C)Y \backslash f(C) is open in YY and hence (Y\f(C))f(X)=f(X)\f(C)(Y \backslash f(C)) \cap f(X) = f(X) \backslash f(C) is open in the subspace topology, which means that f(C)f(C) is closed in the subspace topology.


(projections out of product spaces are open maps)

For (X 1,τ X 1)(X_1,\tau_{X_1}) and (X 2,τ X 2)(X_2,\tau_{X_2}) two topological spaces, then the projection maps

pr i:(X 1×X 2,τ X 1×X 2)(X i,τ X i) 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

X 1×X 2 pr 1 X 1 (x 1,x 2) AAA x 1 \array{ X_1 \times X_2 &\overset{pr_1}{\longrightarrow}& X_1 \\ (x_1, x_2) &\overset{\phantom{AAA}}{\mapsto}& x_1 }
X 1×X 2 pr 2 X 2 (x 1,x 2) AAA x 2 \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 OX 1×X 2O \subset X_1 \times X_2 in the product space topology is a union of products of open subsets U iX 1U_i \in X_1 and V iX 2V_i \in X_2 in the factor spaces

O=iI(U i×V i) 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

pr 1(iI(U i×V i)) =iIpr 1(U i×V i) =iIU i. \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} \,.


A local homeomorphism is an open map.


Let f:XYf \colon X \to Y be a local homeomorphism and UXU \subset X an open subset. We need to see that the image f(U)Yf(U) \subset Y is an open subset of YY. For this we may equivalently show that each yf(U)y \in f(U) has an open neighbourhood inside f(U)f(U).

But since any function is surjective onto its image, there exists xUx \in U with f(x)=yf(x) = y. By local homeomorphy of ff, this xXx \in X has an open neighbourhood U xXU_x \subset X with f |U x:U xf(U x)f_{|U_x} \colon U_x \to f(U_x) a homeomorphism. Since UU xU xU \cap U_x \subset U_x is an open neighbourhood of xx in U xU_x, the homeomorphy of f |U xf_{|U_x} implies that f(UU x)f(U)f(U \cap U_x) \subset f(U) is an open neighbourhood of f(x)=yf(x) = y.



(preimages of open maps preserve topological interiors)
For f:XYf \,\colon\, X \longrightarrow Y an open map and UXU \subset X a subset, the preimage under ff of the interior of UU is the interior of the preimage of all of UU:

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


In one direction, the inclusion

f 1(U )(f 1(U)) 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)Xf^{-1}(U) \subset X by continuity of ff, while the right hand side is the largest such subset, by definition.

In the other direction, the inclusion

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

is equivalent (see there) to the inclusion

f((f 1(U)) )U . 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)f(U) by open-ness of ff, while the right hand side is again the largest open subset of UU, by definition.

And dually:


(preimages of open maps preserve topological closure)
If f:XYf \,\colon\, X \longrightarrow Y is an open map, and UXU \subset X a subset with topological closure U¯\overline{U}, then the preimage f 1(U¯)f^{-1}\big( \overline{U}\big) of the topological closure is the topological closure of the preimage of UU:

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


Noticing that

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

    U¯=X(XU) \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. .

Last revised on June 30, 2022 at 06:39:42. See the history of this page for a list of all contributions to it.