Monotone Light Factorizations

We will show that any continuous map \(f:X\to Y\) of compact Hausdorff spaces can be factored into a composition of two maps where one has connected fibers and the other has totally disconnected fibers. To start, we introduce some terminology and preliminary facts.

Definition: Let \(X\) be a topological space. A decomposition of \(X\) is a partition of \(X\) into compact subsets \(\{C_\alpha\}\). The decomposition \(\{C_\alpha\}\) is upper semi-continuous if whenever a decomposition element \(C_\alpha\) is contained in an open set \(U,\) there exists another open set \(V\subset U\) which also contains \(C_\alpha\) and has the property that for any other \(C_\beta,\) if \(C_\beta\cap V\neq \emptyset\) then \(C_\beta\subset V.\)

The condition to be a USC-decomposition looks a bit strange at first, but we shall see that it is exceedingly convenient and also not too difficult to satisfy in practice.

For \(X\) a topological space and \(\mathcal{D} = \{C_\alpha\}\) an upper semi-continuous decomposition, we write \(X/\mathcal{D}\) for the quotient space where each decomposition element \(C_\alpha\in \mathcal{D}\) is collapsed to a point. As we will see, the condition that \(\mathcal{D}\) is upper semi-continuous is equivalent to \(X/\mathcal{D}\) being Hausdorff.

Lemma: Let \(X\) be a topological space, \(\{C_\alpha\}\) an upper semi-continuous decomposition, and \(U\subset X\) an open set. Then, the union of all the decomposition elements contained in \(U\) is an open set.

Proof: Let \(I = \{\alpha \mid C_\alpha\subset U\}\) and let \(V = \bigcup_{\alpha\in I}C_\alpha.\) Let \(x\in V\) be arbitrary. Then, there is some \(C_\alpha\) such that \(x\in C_\alpha\subset U.\) Because the decomposition \(\{C_\alpha\}\) is USC, there exists an open set \(U'\subset U\) such that \(x\in C_\alpha\subset U'\subset U\) and if \(C_\beta\cap U'\neq \emptyset\) then \(C_\beta\subset U'.\) Let \(y\in U'.\) Since \(\{C_\alpha\}\) is a partition of \(X\), there is a unique \(C_\beta\) containing \(y.\) In particular, this means that \(C_\beta\cap U'\) is non-empty so \(C_\beta\subset U'\subset U,\) i.e. \(y\in V.\) Therefore, \(U'\subset V\) is open and contains \(x.\) Since \(x\in V\) was arbitary, \(V\) is an open set.

Notice that the converse of this Lemma is also true by a very straightforward argument coming from the definition of upper semi-continuity.

Theorem: Let \(X\) be a compact Hausdorff space and let \(\mathcal{D}=\{C_\alpha\}\) be an upper semi-continuous decomposition of \(X\). Then, the quotient space \(X/\mathcal{D}\) is also a compact Hausdorff space.

Proof: The quotient space \(X/\mathcal{D}\) is automatically compact since the quotient map \(p:X\to X/\mathcal{D}\) is surjective and continuous. Let \(C_\alpha,C_\beta\in X/\mathcal{D}.\) Because \(X\) is Hausdorff and \(C_\alpha,C_\beta\subset X\) are compact, they are also closed. Since \(X\) is both compact and Hausdorff, it is also normal so there exists disjoint open sets \(U_\alpha,V_\alpha \subset X\) such that \(C_\alpha\subset U\) and \(C_\beta\subset V.\) Let \(V_\alpha\) be the union of all of the decomposition elements contained in \(U_\alpha\) and let \(V_\beta\) be the union of all the decomposition elements contained in \(U_\beta.\) The sets \(V_\alpha,V_\beta\) are both open by the above lemma. Moreover, \(p(V_\alpha)\) and \(p(V_\beta)\) are open in \(X/\mathcal{D}\) (because of how \(V_\alpha,V_\beta\) were constructed) and are disjoint. Thus, \(X/\mathcal{D}\) is Hausdorff as desired.

Corollary 1: For \(X\) a compact Hausdorff space and \(\mathcal{D} = \{C_\alpha\}\) a decomposition of \(X\), \(\mathcal{D}\) is USC if and only if \(X/\mathcal{D}\) is Hausdorff.

Proof: One direction is exactly the theorem we have just proven, so we just need to show that if \(X/\mathcal{D}\) is Hausdorff, then \(\mathcal{D}\) is USC. Suppose that \(X/\mathcal{D}\) is Hausdorff. Since \(X\) is compact Hausdorff and \(X/\mathcal{D}\) is Hausdorff, the quotient map \(p:X\to X/\mathcal{D}\) is a closed map. Let \(U\subset X\) be open such that \(U\) contains at least one decomposition element. Let \(V\) be the union of all decomposition elements contained in the open set \(U.\) Then, \(p(V)\) is the complement of \(p(X\setminus U)\) so \(p(V)\) is open in \(X/\mathcal{D}.\) Moreover, \(V\) is a union of decomposition elements so \(p^{-1}(p(V))=V\), i.e. \(V\) is also open in \(X\). Thus, \(\mathcal{D}\) is USC.

Corollary 2: Let \(f:X\to Y\) be a surjective continuous map of compact Hausdorff spaces. Then, \(\{f^{-1}(y)\mid y\in Y\}\) is a USC decomposition of \(X\).

Proof: Since \(X,Y\) are compact Hausdorff, \(f\) is a quotient map and every fiber is compact. Moreovoer, \(Y\) being Hausdorff means we can apply Corollary 1 to get that \(\{f^{-1}(y)\mid y\in Y\}\) is USC.

Before tackling the main theorem, we need one more key lemma about USC decompositions which will do most of the work for us in proving the main theorem.

Key Lemma: Let \(X\) be a compact Hausdorff space and let \(\mathcal{D}=\{C_\alpha\}\) be a USC decomposition of \(X.\) Let \(\mathcal{D}'\) be the decomposition whose elements are the connected components of the elements of \(\mathcal{D}.\) Then, \(\mathcal{D}'\) is also USC.

Proof: Let \(C_\alpha'\in \mathcal{D}'\) be a connected component of some \(C_\alpha\in \mathcal{D}\) and suppose that \(C_\alpha'\subset U\) for some open \(U\subset X.\) Because \(C_\alpha\) is compact and Hausdorff, there exists a set \(D\) which is clopen in \(C_\alpha\) such that \(C_\alpha'\subset D\subset (U\cap C_\alpha).\) The set \(D\) is open in \(C_\alpha\) so there exists an open \(U_D\subset X\) such that \(D = U_D\cap C_\alpha.\) Moreover, \(D\) is contained in \(U\) so we can assume that \(U_D\subset U\) (if not just intersect them). Because \(D\) is clopen in \(C_\alpha,\) \(D = \overline{U_D}\cap C_\alpha\) as well. In particular, this means that \((\overline{U_D}\setminus U_D)\cap C_\alpha = \emptyset.\) Define \(U^* = U_D\cup (X\setminus \overline{U_D})\) and note that \(U^*\) is a disjoint union of two open sets and it contains \(C_\alpha.\) Let \(V\) be the union of all the elements in \(\mathcal{D}\) which are contained in \(U^*.\) Since \(\mathcal{D}\) is USC, \(V\) is open in \(X\). Let \(V'=V\cap U_D\) so \(V'\) is open and contains \(C_\alpha'.\) Suppose that \(C_\beta'\) is a component of some \(C_\beta\in \mathcal{D}\) such that \(C_\beta\) intersects \(V'\) non-trivially. By construction, \(C_\beta\) must be contained within \(V.\) Moreover, \(C_\beta'\) is connected and intersects with \(U_D\) so \(C_\beta'\) must be contained in \(U_D.\) Thus, \(C_\beta'\subset V'\subset U_D\subset U.\) The open set \(V'\) therefore demonstrates that \(\mathcal{D}'\) is USC.

Definition: A continuous map \(f:X\to Y\) is monotone if all of its fibers are connected.

Definition: A continuous map \(f:X\to Y\) is light if all of its fibers are totally disconnected.

We now have all of the ingredients we need to state and prove the Monotone-Light Factorization theorem:

Theorem (Monotone-Light Factorization): Let \(f:X\to Y\) be a continuous mapping of compact Hausdorff spaces. Then, there exists a third compact Hausdorff space \(Z\) and continuous maps \(m:X\to Z\) and \(l:Z\to Y\) such that \(f = l\circ m\), \(m\) is monotone, and \(l\) is light.

Proof: Let \(\mathcal{D}\) be the decomposition of \(X\) given by the connected components of the fibers of \(f.\) Corollary 2 and our Key Lemma guarantee that this is a USC decomposition of \(X\). Let \(Z\) be the quotient space \(X/\mathcal{D}\) and let \(m:X\to Z\) be the quotient map. The space \(Z\) is compact Hausdorff by our previous theorem and \(m\) is monotone by construction. There is a natural map \(l:Z\to Y\) which just takes a component of the fiber \(f^{-1}(y)\) to the point \(y.\) The map \(l\) is light because the collection of components of a fiber \(f^{-1}(y)\) will be totally disconnected in the quotient space \(Z = X/\mathcal{D}.\)

References:

Danny Calegari’s Point Set-Topology Course at UChicago Winter 2023.

Decompositions of Manifolds by Robert J. Daverman.