# Preorder on topological spaces whose equivalence classes are the homeomorphism classses

### Problem

Define a preorder $$\preceq$$ on topological spaces such that

A) $$X \preceq Y$$ and $$Y \preceq X$$ implies $$X$$ and $$Y$$ are homeomorphic.

This preorder should not be the trivial solution of "X and Y are homeomorphic".

### Motivation

I'm interested in whether there is a "directed" (for a lack of a better word) version of a homeomorphism.

### Failing candidates

Some of the candidates I thought of were:

• Continuous bijections. Define $$X \preceq Y$$ if there is a continuous bijection from $$X$$ to $$Y$$. Property A fails.

• Surjective continuous closed maps. Define $$X \preceq Y$$ if there is a surjective continuous closed map from $$X$$ to $$Y$$. That property A fails can be seen by considering a space-filling curve $$f: [0, 1] \to [0, 1]^2$$ and the projection $$g: [0, 1]^2 \to [0, 1]$$. The projection here is closed because of compactness, and the space-filling curve is closed as a continuous map from compact space to Hausdorff space by the closed map lemma.

• Quotient maps. Define $$X \preceq Y$$ if there is a quotient map from $$X$$ to $$Y$$. Previous item shows that this also fails property A, since surjective continuous closed maps are quotient maps.

• Embeddings. Define $$X \preceq Y$$ if there is an embedding from $$X$$ to $$Y$$. Property A fails.

• Covering maps. Define $$X \preceq Y$$ if there is a covering map from $$X$$ to $$Y$$. Property A fails.

• Surjective local homeomorphisms. Define $$X \preceq Y$$ if there is a surjective local homeomorphism from $$X$$ to $$Y$$. A covering map is a surjective local homeomorphism. By the previous item, property A fails.

• Surjective continuous open maps. Define $$X \preceq Y$$ if there is a surjective continuous open map from $$X$$ to $$Y$$. A surjective local homeomorphism is a surjective continuous open map. By the previous item, property A fails.

### Unclear candidates

• Hereditary quotient maps. Define $$X \preceq Y$$ if there is a hereditary quotient map from $$X$$ to $$Y$$. I don't know whether this works or not. A function $$f: X \to Y$$ is a hereditary quotient map, if $$f|X': X' \to Y'$$ is a quotient map for each $$Y' \subset Y$$ and $$X' = f^{-1}[Y']$$.
• I don't think that there will be a good answer to this problem. Maybe for some special classes of topological spaces. Dec 1, 2023 at 23:21
• Of course there are other ways to cheat. Say $X\preceq Y$ if $|X|<|Y|$ or $X,Y$ are homeomorphic. Dec 1, 2023 at 23:23
• Maybe what you actually want is a subcategory $\mathcal{C}$ of $\mathbf{Top}$ containing all objects (that is, we really just have special types of continuous maps, and these compose) such that (a) $\mathcal{C}$ contains all isomorphisms, (b) $\mathcal{C}$ contains non-isomorphisms, (c) if there are morphisms $X \to Y$ and $Y \to X$ in $\mathcal{C}$, then $X \cong Y$. Dec 1, 2023 at 23:26
• @StevenClontz Added counter-example for embeddings.
– kaba
Dec 1, 2023 at 23:40
• I think the most natural "trivial" example is "either $X$ and $Y$ are homeomorphic, or else $X$ is embeddale in $Y$ but $Y$ is not embeddable in $X$."
– bof
Dec 2, 2023 at 0:19