Let $\kappa$ be a cardinal, and let $\text{Top}(\kappa)$ be the set of topological spaces $(X,\tau)$ such that $X\subseteq \kappa$. We pre-order $\text{Top}(\kappa)$ by

for $X, Y \in \text{Top}(\kappa)$ we set $X \to Y$ iff there is a continuous surjective map $f:X\to Y$.

We say $X\simeq Y$ iff $X\to Y$ and $Y\to X$. So the relation $\geq$ on $\text{Top}(\kappa)/\simeq$ defined by $[X]_\simeq \geq [Y]_\simeq$ iff $X\to Y$ is well-defined, and $(\text{Top}(\kappa)/\simeq, \geq)$ is a poset.

Is $(\text{Top}(\kappa)/\simeq, \geq)$ also a lattice? Is it complete? Has it been studied anywhere?

  • $\begingroup$ I am not sure if the empty set is considered a topological space, but in the context of this question it certainly should not be. $\endgroup$ – Goldstern Mar 25 '15 at 11:01
  • $\begingroup$ The Rudin-Keisler ordering can be formulated in terms of continuous mappings between spaces. In particular, if $X,Y$ are discrete spaces, then each mapping $f:X\rightarrow Y$ extends uniquely to a continuous map $\overline{f}:\beta X\rightarrow\beta Y$ between the Stone-Cech compactifications. If $u\in\beta X,v\in\beta Y$, then $v\leq_{RK}u$ iff there is a continuous mapping $f:X\rightarrow Y$ such that $\overline{f}(u)=v$. Similarly, the Wadge heirarchy is another notion of reducibility by continuous mappings. $\endgroup$ – Joseph Van Name Mar 25 '15 at 18:58

A partial answer: It is not a lattice, at least if you allow non-Hausdorff spaces.

For any natural numbers $n$ and $k$, write $\bf n+k$ for the sum of an indiscrete space of size $n$ (i.e., a space that has no nontrivial open sets) and an indiscrete space of size $k$. Then both $\bf 4$ and $\bf 2+1$ are lower bounds for the set $\{\bf 2+2, \, 3+1 \}$, but there is no greatest lower bound.

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I have seen this partial order before (for a specific context) in Hart J.E., Kunen K., Bohr compactifications of discrete structures, Fund. Math. 160 (1999). Let me briefly explain the situation considered there:

Fix a set $A$. A pair $(X,\varphi)$ is a compactification of $A$ if $X$ is a compact Hausdorff space and $\varphi:A \to X$ is a (not necessarily injective) function such that $\varphi(A)$ is dense in $X$. Define $(X,\varphi) \leq (Y,\psi)$ if there exists a continuous $f:Y \to X$ such that $f \circ \psi=\varphi$. Note that the last condition implies that $f$ is surjective since $f(\psi(A))$ is dense in $X$.

Now just as you did, they define the equivalence relation and the partial order on the set $\mathbb{K}(A)$ of all equivalence classes: $(X,\varphi) \sim (Y,\psi)$ if and only if $(X,\varphi) \leq (Y,\psi)$ and $ (Y,\psi) \leq (X,\varphi)$; $[(X,\varphi)] \leq [(Y,\psi)]$ if and only if $(X,\varphi) \leq (Y,\psi)$. The following is Lemma 2.2.7:

$\mathbb{K}(A)$ is a complete lattice.

You can also add a topology and/or a first order structure to the set $A$ and study a restricted version of compactifications. For instance this gives a way of defining the Bohr compactification of a topological group.

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