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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 $X' = f^{-1}[Y']$.
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  • $\begingroup$ I don't think that there will be a good answer to this problem. Maybe for some special classes of topological spaces. $\endgroup$ Dec 1, 2023 at 23:21
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    $\begingroup$ Of course there are other ways to cheat. Say $X\preceq Y$ if $|X|<|Y|$ or $X,Y$ are homeomorphic. $\endgroup$ Dec 1, 2023 at 23:23
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    $\begingroup$ 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 $X \cong Y$. $\endgroup$ Dec 1, 2023 at 23:26
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    $\begingroup$ @StevenClontz Added counter-example for embeddings. $\endgroup$
    – kaba
    Dec 1, 2023 at 23:40
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    $\begingroup$ 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 $X$." $\endgroup$
    – bof
    Dec 2, 2023 at 0:19

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