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Assume all spaces are Hausdorff.

I am looking for a characterization of spaces $X$ such that if $f:X\to Y$ is a continuous one-to-one surjection then $f$ is a homeomorphism. I know this is true when $X$ is compact.

Is compactness necessary?


POSSIBLE SOLUTION:

Here we will work with a separable metric space $X$, so that $X$ has a metric compactification $\gamma X$ (for instance, the closure of an embedding in the Hilbert cube).

Supposing that $X$ is not compact, we will construct a continuous one-to-one mapping that is not a homeomorphism.

Let $x\in X$. Since $X$ is not compact, there exists $p\in \gamma X\setminus X$. Let $\varphi:\gamma X\to \gamma X/\{p,x\}$ be the canonical epimorphism onto the quotient obtained by identifying $p$ and $x$. Let $f=\varphi\restriction X$ and $Y=\varphi[X]$. Then $f:X\to Y$ is continuous and one-to-one, but is not a closed mapping. To see this, note that there is a sequence $(x_i)_{i\in\omega}$ of points in $X\setminus \{x\}$ converging to $p$ because $X$ is dense in $\gamma X$. Then $A:=\{x_i:i\in\omega\}$ is closed in $X$, but $x\in \overline{f[A]}\setminus f[A]$. $\square$


Notes: I think this approach would work if $X$ simply has a compactification. The compactification probably does not even have to be Hausdorff, so $X$ just needs to be Hausdorff I think. Does that sound right? I don't know how it would go when you have lesser separation axioms. Like is there a $T_1$ example that is not compact?

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  • $\begingroup$ Is this true for discrete spaces? If $X$ is any topological space, and $X^\delta$ is the set $X$ with the discrete topology, then the identity map in the category of sets $id_X:X^\delta\to X$ is a continuous bijection but is not a homeomorphism $\endgroup$ – freeRmodule Dec 5 '16 at 1:39
  • $\begingroup$ Oh right... my brain is not working today $\endgroup$ – Forever Mozart Dec 5 '16 at 1:41
  • $\begingroup$ Indiscrete topology is another. $\endgroup$ – Brian M. Scott Dec 5 '16 at 1:44
  • $\begingroup$ This is not true for $X$ compact Hausdorff unless you require $Y$ to be Hausdorff as well. $\endgroup$ – Eric Wofsey Dec 5 '16 at 3:42
  • $\begingroup$ correct. let's assume everything is Hausdorff, or even metric if you like $\endgroup$ – Forever Mozart Dec 5 '16 at 5:04

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