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Suppose that $(X,\tau_X)$ and $(Y,\tau_Y)$ are two topological spaces where neither is given the discrete (or indiscrete) topology.

  • Does there always exists a nonconstant function $f:X\to Y$ is such that for all $Z\subseteq X$ if $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq f(\operatorname{Cl}_X(Z))$ then $f^{-1}(\operatorname{Cl}_Y{\left(f(Z)\right)})\subseteq \operatorname{Cl}_X{\left(Z\right)}$?

  • If not, then what properties are necessary and sufficient for the topological spaces to ensure the existence of such a function?

We note that if either $f$ is a closed continuous map or is injective then $f$ satisfies the condition. So if the $X$ and $Y$ has at least the same cardinality then the existence of $f$ is guaranteed. The remaining question thus is the following,

Suppose that $(X,\tau_X)$ and $(Y,\tau_Y)$ are two topological spaces where neither is given the discrete (or indiscrete) topology and such that the cardiality of $Y$ is strictly less that that of $X$.

  • Does there always exists a nonconstant function $f:X\to Y$ is such that for all $Z\subseteq X$ if $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq f(\operatorname{Cl}_X(Z))$ then $f^{-1}(\operatorname{Cl}_Y{\left(f(Z)\right)})\subseteq \operatorname{Cl}_X{\left(Z\right)}$?

  • If not, then what properties are necessary and sufficient for the topological spaces to ensure the existence of such a function?

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    $\begingroup$ Why "if $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq \operatorname{Cl}_Y(f(Z))$" ? That's as restrictive as "if $1+1=2$" ... $\endgroup$ Apr 9, 2020 at 5:28
  • $\begingroup$ @HagenvonEitzen: Nice catch. Thanks. $\endgroup$
    – user170039
    Apr 9, 2020 at 5:36

1 Answer 1

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If both $X$ and $Y$ are $T_1$ spaces then any function $f$ satisfying the condition is injective. Indeed, let $x\in X$ be any point. Put $Z=\{x\}$. Then $\operatorname{Cl}_X(Z)=\{x\}$, $\operatorname{Cl}_Y(f(Z))=\{f(x)\}$, thus $\operatorname{Cl}_Y{\left(f(Z)\right)}\subseteq f(\operatorname{Cl}_X(Z))$. Thus $$ f^{-1}(f(x))=f^{-1}(\operatorname{Cl}_Y{\left(f(Z)\right)})\subseteq \operatorname{Cl}_X{\left(Z\right)}= \{x\}.$$

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