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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 $X' = f^{-1}[Y']$.
kaba
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  • I don't think that there will be a good answer to this problem. Maybe for some special classes of topological spaces. – Martin Brandenburg Dec 01 '23 at 23:21
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    Of course there are other ways to cheat. Say $X\preceq Y$ if $|X|<|Y|$ or $X,Y$ are homeomorphic. – Steven Clontz Dec 01 '23 at 23:23
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    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 $X \cong Y$. – Martin Brandenburg Dec 01 '23 at 23:26
  • I don't think this works but don't have a counter example: $X\preceq Y$ means X embeds in Y. – Steven Clontz Dec 01 '23 at 23:33
  • Clearly @StevenClontz comment is an answer to the question. But like he says, it's somehow "cheating" even though I cannot put my finger on why that is. I'm hoping here that people get the spirit of the question though. There's some assumption that is missing to make the question more interesting. – kaba Dec 01 '23 at 23:36
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    @StevenClontz Added counter-example for embeddings. – kaba Dec 01 '23 at 23:40
  • I think the point is that $X \preceq Y$ should preferably be some widely useful fact / well-known commonly used tool, such as embeddability (which does not work). As for @StevenClontz cardinality trick, it is not generally useful to know that $|X| < |Y|$ or $X$ and $Y$ are homeomorphic (or who knows, maybe there is some application :). – kaba Dec 01 '23 at 23:49
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    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 02 '23 at 00:19
  • There's a very easy counter example to my embedding pitch: $(-1,1)\subseteq[-1,1]\subseteq(-2,2)$. – Steven Clontz Dec 02 '23 at 01:21

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