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Definition. Let $S$ and $T$ be topological spaces and $f:S \to T$ . $f$ is a homeomorphism is $f$ is a one-to-one correspondence and both $f$ and $f^{-1}$ are continuous . (Here one-to-one correspondence means both injective and surjective)

Since for real-valued function in $R$ , $f$ is a one-to-one correspondence and continuous implies $f^{-1}$ is continuous . Does this still hold for arbitary topological space $S$ and $T$ ? Or , if $f:S \to T$ . $f$ is a one-to-one correspondence continuous function . $V $ is an open subset of $S$ , can we prove that $f(V)$ is an open subset of $T$?

J.Guo
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3 Answers3

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In general no, take $f:\{0\}\to\Bbb{R}$ to be $f(0)=0$. This map is injective and continuous ($\Bbb{R}$ with the usual topology; $\{0\}$ with the subspace topology).

However if $S$ is assumed to be an open subset of $\Bbb{R}^n$ and $T=\Bbb{R}^n$, then $f$ is an open map, see invariance of domain

(Edit: here is an example of $f$ being bijective and continuous but $f^{-1}$ is not continuous:

Take $f:[0,1)\cup [2,3]\to [0,2]$ (with the usual subspace topology of $\Bbb{R}$) be $$f(x)=\begin{cases} x &, x\in[0,1)\\ x-1 &, x\in [2,3] \end{cases}$$ I suggest you draw the graphs of $f$ and $f^{-1}$ so that it will be easier to see that $f$ is continuous and $f^{-1}$ is not.)

Nick
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  • I think here $f$ is not surjective , so it is not a one-to-one correspondence . – J.Guo Dec 29 '18 at 06:27
  • @J.Guo I see, just put an example in. – Nick Dec 29 '18 at 06:37
  • @J.Guo take $[2,3]$ in the domain which is open, but $[1,2]$ isn't open in the range since $1$ is not in the interior of the set $[1,2]$(open=every point of the set is an interior point). – Manraj Dec 29 '18 at 07:07
  • I see , thank you ! – J.Guo Dec 29 '18 at 07:11
  • @RakeshBhatt I think you know how to read the last time I edited my answer and the time for yours. I dont think I have the power to copy your example before you posted it. – Nick Dec 29 '18 at 10:45
  • Weird, The time when I posted it you post was having 1st Para. Sorry :/ – roasted_cashews Dec 29 '18 at 10:54
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Here another example could be $f:[0,2\pi)\rightarrow S_1$ defined by $f=e^{i\theta}$ with both sets with subspace topology. This is both bijective and continuous(you can check). Now, $[0,\pi)$ is open in $[0,2\pi)$ but $f[0,\pi)$ isn't open in $S_1$ because $0$ isn't in the interior of the set.

Manraj
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Consider the function $$ f: [0,1) \cup [2,3] \to [0,2], \; f(x) = \begin{cases}x , & x \in [0,1) \\ x-1, & x \in [2,3] \end{cases}. $$

This is a continuous bijection. The set $[2,3]$ is open in $[0,1) \cup [2,3]$. But the image of $[2,3]$ is $[1,2]$ which is not open in $[0,2]$.

roasted_cashews
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