I suppose the question is in the title. My immediate thought would be to extend to set-valued functions, and that doesn't seem like a huge problem, at least not obviously. For example, you could "blow up" (abuse of terminology) a point into an open set of a separate topology by mapping that point onto an interval.

Does this idea make sense, and if so, where can I find references to it, if any?

EDIT: Noah's comment has given some axioms for some weak relations satisfying these properties:

Given topological spaces $(X,\tau)$, $(Y,\sigma)$, we say a relation $R \subseteq X\times Y$ is almost homeomorphic if it satisfies the following axioms:

$(i)$ For each $U \in \tau$ we have $R(U)\in \sigma$.

$(ii)$ For each $V \in \sigma$, we have $R^{-1}(V)\in \tau$.

$(iii)$ For each $x \in X$, there is at least one $y \in Y$ such that $xRy$ is true.

  • 1
    $\begingroup$ @aldodecristo In general you do get not local homeomorphisms. Consider any map $\mathbb R \to \{*\}$. $\endgroup$
    – Paul Frost
    Mar 4, 2021 at 17:30
  • $\begingroup$ Do you mean if we assume that maps are continuous and open? $\endgroup$
    – Randall
    Mar 4, 2021 at 17:32
  • $\begingroup$ "an embedding is a homeomorphism onto its image", en.wikipedia.org/wiki/Embedding#General_topology $\endgroup$
    – lhf
    Mar 4, 2021 at 18:21
  • 1
    $\begingroup$ Per your comment about blowing up points, it sounds like you're interested in the following: given topological spaces $(X,\tau),(Y,\sigma)$, a relation $R\subseteq X\times Y$ such that $(i)$ for each $U\in\tau$ we have $R(U)\in\sigma$, $(ii)$ for each $V\in\sigma$ we have $R^{-1}(U)\in\tau$, and $(iii)$ for each $x\in X$ there is at least one $y\in Y$ with $xRy$. So we drop a priori injectiveness, surjectiveness, and single-valuedness. Does that seem right? $\endgroup$ Mar 4, 2021 at 18:55
  • 2
    $\begingroup$ Any map $f:(X,\tau_X)\rightarrow(Y,\tau_Y)$ induces a function $f^*:\tau_Y\rightarrow \tau_X$, $U\mapsto f^{-1}(U)$. I have seen the definition that $f$ is a quasihomeomorphism if $f^*$ is bijective. Obviously every homeomorphism is a quasihomeomorphism, but there are quasihomeomorphisms which are not bijective. If $X$ is any space and $X_0$ the quotient $T_0$ space formed by identifying $x\sim y$ iff $\overline{\{x\}}=\overline{\{y\}}$, then $X\rightarrow X_0$ is a surjective quasihomeomorphism which is bijective iff $X$ is $T_0$. $\endgroup$
    – Tyrone
    Mar 4, 2021 at 19:18


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