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

The basic type of function considered in topology is a continuous function between spaces. $f$ is **continuous** if for every open subset $U$ of $Y$ the set $f^{-1}(U)$ is an open subset of $X$. 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.

An open map is in some sense the dual notion to a continuous function, $f$ is an **open map** if $V$ is an open subset of $X$ implies $f(V)$ is an open subset of $Y$. 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.

A closed map is similar in definition to an open map. $f$ is a **closed map** if $C$ being a closed subset of $X$ implies that $f(C)$ is a closed subset of $Y$. 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 and closed maps are distinct from each other! For an example, take $\pi: \mathbb{R}^2 \rightarrow \mathbb{R}$ which projects the plane onto the $x$-axis. $\pi$ is in fact an open map, but we can realize a closed set whose image is open: namely the graph of $y=1/x$ whose image is the complement of $\{ 0 \}$ 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 $g: (\mathbb{N}_{\geq 0},\tau) \rightarrow (\mathbb{N}_{\geq 0},\tau')$ be defined by $g(n)=n+1$. We must first define the topologies $\tau$ and $\tau'$ and it will then become trivial to show that $f$ is a closed continuous function, but not open. Let $\tau$ the the *cofinite topology* that is a subset is open if it’s complement is finite. Let $\tau'$ be the intersection of $\tau$ and the set of subsets containing $0$, so that a set is open if it contains $0$ 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 $0$. Clearly $g$ sends closed sets under $\tau$ to closed sets under $\tau'$, but if $U$ is a $\tau$ open set, then $0 \notin g(U)$ so that $g(U)$ is not open. It is true here that $g$ is continuous, but we will not prove it.

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

That is: why isn’t there a condition such as $f^{-1}(C)$ is closed for every closed $C$ 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