Joyal's CatLab Functions between topological spaces

Functions between topological spaces

For the rest of this page, we define f:XYf: X \rightarrow Y a function between two topological spaces.

Continuous functions

The basic type of function considered in topology is a continuous function between spaces. ff is continuous if for every open subset UU of YY the set f 1(U)f^{-1}(U) is an open subset of XX. It becomes a simple exercise to show that a composition of continuous functions is continuous.

Often when one is dealing very regularly with continuous functions, the term map is used as a shorthand for continuous function.

Open maps

An open map is in some sense the dual notion to a continuous function, ff is an open map if VV is an open subset of XX implies f(V)f(V) is an open subset of YY. Note that even in contexts where map means continuous function, an open map is generally not required to be continuous. It is again a simple exercise to show composition of open maps is an open map.

Closed maps

A closed map is similar in definition to an open map. ff is a closed map if CC being a closed subset of XX implies that f(C)f(C) is a closed subset of YY. Again we have that showing a composition of closed maps is again a closed map as an exercise, and the same warning from above applies. Closed maps are generally not required to be continuous, even when the context is such that a map is a continuous function.

Open maps are distinct from closed maps

Open maps and closed maps are distinct from each other! For an example, take π: 2\pi: \mathbb{R}^2 \rightarrow \mathbb{R} which projects the plane onto the xx-axis. π\pi is in fact an open map, but we can realize a closed set whose image is open: namely the graph of y=1/xy=1/x whose image is the complement of {0}\{ 0 \} as a subset of \mathbb{R}.

Conversely there are closed maps which are not open, I will construct one here, but it is somewhat contrived(so please bear with me!). Let g:( 0,τ)( 0,τ)g: (\mathbb{N}_{\geq 0},\tau) \rightarrow (\mathbb{N}_{\geq 0},\tau') be defined by g(n)=n+1g(n)=n+1. We must first define the topologies τ\tau and τ\tau' and it will then become trivial to show that ff is a closed continuous function, but not open. Let τ\tau the the cofinite topology that is a subset is open if it’s complement is finite. Let τ\tau' be the intersection of τ\tau and the set of subsets containing 00, so that a set is open if it contains 00 and it’s complement is finite. The closed subsets under the former topology are any finite subset, under the latter they are finite sets not containing 00. Clearly gg sends closed sets under τ\tau to closed sets under τ\tau', but if UU is a τ\tau open set, then 0g(U)0 \notin g(U) so that g(U)g(U) is not open. It is true here that gg is continuous, but we will not prove it.

However, if ff is bijective, it is a closed map if and only if it is a open map.

Why isn’t there something like a closed-continuous function?

That is: why isn’t there a condition such as f 1(C)f^{-1}(C) is closed for every closed CC in the codomain? This is actually the same condition as continuity, as complements commute with function preimages of set functions.


A homeomorphism is a continuous bijection whose set function inverse is also continuous, that is a bijective continuous open map.

Created on September 1, 2012 at 23:48:05 by jstalfos