Open Mappings and Closed Mappings

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Definitions and Basic Properties

Definition 1 For any two topological spaces \(X,Y\) and a function \(f:X \rightarrow Y\), we define the following.

1. If for any open set \(U\) of \(X\), \(f(U)\) is always an open set of \(Y\), then we call \(f\) an open mapping.
2. If for any closed set \(A\) of \(X\), \(f(A)\) is always a closed set of \(Y\), then we call \(f\) a closed mapping.

Then the following holds.

Proposition 2 Let topological spaces \(X,Y,Z\) and functions \(f:X \rightarrow Y\), \(g:Y \rightarrow Z\) be given. Then the following holds.

1. If \(f\) and \(g\) are both open (resp. closed), then \(g\circ f\) is also open (resp. closed).
2. If \(g\circ f\) is open (resp. closed) and \(f\) is a continuous surjection, then \(g\) is open (resp. closed).
3. If \(g\circ f\) is open (resp. closed) and \(g\) is a continuous injection, then \(f\) is open (resp. closed).

Proof

1. Obvious.

2. Let \(V\) be an arbitrary open set of \(Y\). Then since \(f\) is continuous, \(f^{-1}(V)\) is an open set of \(X\), and thus

   \[(g\circ f)(f^{-1}(V))=g(f(f^{-1}(V)))=g(V)\]

   holds, so \(g(V)\) is an open set of \(Z\). Here we used that \(f(f^{-1}(V))=V\) since \(f\) is a surjection. ([Set Theory] §Retraction and Section, ⁋Definition 2) On the other hand, if \(B\) is an arbitrary closed set of \(Y\), then by §Continuous Functions, ⁋Theorem 4 we can apply the same argument as above.

3. As in the second proof, it suffices to consider the open case. For any open set \(U\) of \(X\), since \(g\circ f\) is open, \(g(f(U))\) is open, and thus using that \(g\) is continuous and the above [Set Theory] §Retraction and Section, ⁋Definition 2, from the following formula

   \[g^{-1}(g(f(U))=f(U)\]

   we see that \(f(U)\) is an open set.

The following proposition helps determine when a function between topological spaces is open or closed.

Proposition 3 Let a function \(f:X \rightarrow Y\) between two topological spaces be given. Then the following holds.

1. If \(f\) is open (resp. closed), then for any subset \(A\) of \(Y\), the restriction \(f\vert_{f^{-1}(A)}: f^{-1}(A) \rightarrow A\) is also open (resp. closed).
2. Let \((A_i)_{i\in I}\) be a covering of \(Y\) that is either (1) a locally finite closed covering, or (2) \((\interior A_i)_{i\in I}\) is an open covering of \(Y\). If each restriction \(f\vert_{f^{-1}(A_i)}\) is open (resp. closed), then so is \(f\).

Proof

To show the first result, choose an open set (resp. closed set) in \(f^{-1}(A)\). Then there exists an open set (resp. closed set) \(U\) in \(X\) such that we can write it in the form \(U\cap f^{-1}(A)\). Therefore

\[f\vert_{f^{-1}(A)}(U)=f(U)\cap A\]

holds, and by assumption \(f(U)\) is an open set (resp. closed set), so we obtain the desired result.

The second result can be proved similarly. Let \(U\) be an open set (resp. closed set) in \(X\), and define \(U_i\) by the following formula

\[U_i=U\cap f^{-1}(A_i)\]

Then \(f\vert_{f^{-1}(A_i)}(U_i)=f(U)\cap A_i\) holds, and thus by assumption \(f(U)\cap A_i\) is an open set (resp. closed set) for all \(i\). Therefore, if \(U\) is an open set then \(f(U)\) is a union of open sets and hence open, and if \(U\) is a closed set then by §Interior, Closure, and Boundary, ⁋Proposition 4 it is a closed set.

Equivalence Relations

Definition 4 An equivalence relation \(R\) defined on a topological space \(X\) is said to be open (resp. closed) if the canonical map \(X \rightarrow X/R\) is open (resp. closed).

Then the following can be easily shown.

Proposition 5 Consider a continuous function \(f:X \rightarrow Y\) between two topological spaces and its canonical decomposition

\[X \overset{p}{\longrightarrow} X/R \overset{h}{\longrightarrow h} f(X)\overset{i}{\longrightarrow}Y\]

Then the following are all equivalent.

1. \(f\) is open (resp. closed).
2. \(p\), \(h\), and \(i\) are all open (resp. closed).
3. \(R\) is open (resp. closed), \(h\) is a homeomorphism, and \(f(X)\) is an open (resp. closed) subset of \(Y\).

Properties of Open Mappings

Now we examine the properties that open mappings and closed mappings respectively possess. We begin with the case of open mappings.

Proposition 6