Does there always exist a continuous map saturating a given open set?












5












$begingroup$


Let $X$ and $Y$ be two general topological spaces. Is the following statement true?



For any open $Usubset X$, there exists an open $Vsubset Y$ and a continuous map $f:Xrightarrow Y$, such that $f^{-1}V=U$.






general-topology continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
    $endgroup$
    – copper.hat
    Jan 16 at 14:57










  • $begingroup$
    But there are potentially many $f$'s?
    $endgroup$
    – Uchiha
    Jan 16 at 14:59










  • $begingroup$
    You are correct, sloppy thinking on my part.
    $endgroup$
    – copper.hat
    Jan 16 at 15:00
















5












$begingroup$


Let $X$ and $Y$ be two general topological spaces. Is the following statement true?



For any open $Usubset X$, there exists an open $Vsubset Y$ and a continuous map $f:Xrightarrow Y$, such that $f^{-1}V=U$.






general-topology continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
    $endgroup$
    – copper.hat
    Jan 16 at 14:57










  • $begingroup$
    But there are potentially many $f$'s?
    $endgroup$
    – Uchiha
    Jan 16 at 14:59










  • $begingroup$
    You are correct, sloppy thinking on my part.
    $endgroup$
    – copper.hat
    Jan 16 at 15:00














5












5








5


1



$begingroup$


Let $X$ and $Y$ be two general topological spaces. Is the following statement true?



For any open $Usubset X$, there exists an open $Vsubset Y$ and a continuous map $f:Xrightarrow Y$, such that $f^{-1}V=U$.






general-topology continuity







share|cite|improve this question









$endgroup$




Let $X$ and $Y$ be two general topological spaces. Is the following statement true?



For any open $Usubset X$, there exists an open $Vsubset Y$ and a continuous map $f:Xrightarrow Y$, such that $f^{-1}V=U$.






general-topology continuity




general-topology continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 14:53









UchihaUchiha

12211




12211












  • $begingroup$
    Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
    $endgroup$
    – copper.hat
    Jan 16 at 14:57










  • $begingroup$
    But there are potentially many $f$'s?
    $endgroup$
    – Uchiha
    Jan 16 at 14:59










  • $begingroup$
    You are correct, sloppy thinking on my part.
    $endgroup$
    – copper.hat
    Jan 16 at 15:00


















  • $begingroup$
    Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
    $endgroup$
    – copper.hat
    Jan 16 at 14:57










  • $begingroup$
    But there are potentially many $f$'s?
    $endgroup$
    – Uchiha
    Jan 16 at 14:59










  • $begingroup$
    You are correct, sloppy thinking on my part.
    $endgroup$
    – copper.hat
    Jan 16 at 15:00
















$begingroup$
Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
$endgroup$
– copper.hat
Jan 16 at 14:57




$begingroup$
Take $Y$ to be finite, and $X$ to have an infinite number of open sets.
$endgroup$
– copper.hat
Jan 16 at 14:57












$begingroup$
But there are potentially many $f$'s?
$endgroup$
– Uchiha
Jan 16 at 14:59




$begingroup$
But there are potentially many $f$'s?
$endgroup$
– Uchiha
Jan 16 at 14:59












$begingroup$
You are correct, sloppy thinking on my part.
$endgroup$
– copper.hat
Jan 16 at 15:00




$begingroup$
You are correct, sloppy thinking on my part.
$endgroup$
– copper.hat
Jan 16 at 15:00










1 Answer
1






active

oldest

votes


















3












$begingroup$

If $Y$ only has one element then $f^{-1}(V)in{varnothing,X}$ for any open set $V$.



So if $X$ has a non-trivial open set then it does not work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $f^{-1}(V)={emptyset,X}$?
    $endgroup$
    – SmileyCraft
    Jan 16 at 15:03






  • 2




    $begingroup$
    @SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
    $endgroup$
    – drhab
    Jan 16 at 15:05












  • $begingroup$
    Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
    $endgroup$
    – Uchiha
    Jan 16 at 15:08










  • $begingroup$
    Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
    $endgroup$
    – drhab
    Jan 16 at 15:14












  • $begingroup$
    Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
    $endgroup$
    – Uchiha
    Jan 16 at 15:25














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

If $Y$ only has one element then $f^{-1}(V)in{varnothing,X}$ for any open set $V$.



So if $X$ has a non-trivial open set then it does not work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $f^{-1}(V)={emptyset,X}$?
    $endgroup$
    – SmileyCraft
    Jan 16 at 15:03






  • 2




    $begingroup$
    @SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
    $endgroup$
    – drhab
    Jan 16 at 15:05












  • $begingroup$
    Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
    $endgroup$
    – Uchiha
    Jan 16 at 15:08










  • $begingroup$
    Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
    $endgroup$
    – drhab
    Jan 16 at 15:14












  • $begingroup$
    Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
    $endgroup$
    – Uchiha
    Jan 16 at 15:25


















3












$begingroup$

If $Y$ only has one element then $f^{-1}(V)in{varnothing,X}$ for any open set $V$.



So if $X$ has a non-trivial open set then it does not work.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $f^{-1}(V)={emptyset,X}$?
    $endgroup$
    – SmileyCraft
    Jan 16 at 15:03






  • 2




    $begingroup$
    @SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
    $endgroup$
    – drhab
    Jan 16 at 15:05












  • $begingroup$
    Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
    $endgroup$
    – Uchiha
    Jan 16 at 15:08










  • $begingroup$
    Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
    $endgroup$
    – drhab
    Jan 16 at 15:14












  • $begingroup$
    Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
    $endgroup$
    – Uchiha
    Jan 16 at 15:25
















3












3








3





$begingroup$

If $Y$ only has one element then $f^{-1}(V)in{varnothing,X}$ for any open set $V$.



So if $X$ has a non-trivial open set then it does not work.






share|cite|improve this answer











$endgroup$



If $Y$ only has one element then $f^{-1}(V)in{varnothing,X}$ for any open set $V$.



So if $X$ has a non-trivial open set then it does not work.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 15:05

























answered Jan 16 at 15:02









drhabdrhab

104k545136




104k545136












  • $begingroup$
    Why is $f^{-1}(V)={emptyset,X}$?
    $endgroup$
    – SmileyCraft
    Jan 16 at 15:03






  • 2




    $begingroup$
    @SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
    $endgroup$
    – drhab
    Jan 16 at 15:05












  • $begingroup$
    Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
    $endgroup$
    – Uchiha
    Jan 16 at 15:08










  • $begingroup$
    Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
    $endgroup$
    – drhab
    Jan 16 at 15:14












  • $begingroup$
    Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
    $endgroup$
    – Uchiha
    Jan 16 at 15:25




















  • $begingroup$
    Why is $f^{-1}(V)={emptyset,X}$?
    $endgroup$
    – SmileyCraft
    Jan 16 at 15:03






  • 2




    $begingroup$
    @SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
    $endgroup$
    – drhab
    Jan 16 at 15:05












  • $begingroup$
    Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
    $endgroup$
    – Uchiha
    Jan 16 at 15:08










  • $begingroup$
    Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
    $endgroup$
    – drhab
    Jan 16 at 15:14












  • $begingroup$
    Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
    $endgroup$
    – Uchiha
    Jan 16 at 15:25


















$begingroup$
Why is $f^{-1}(V)={emptyset,X}$?
$endgroup$
– SmileyCraft
Jan 16 at 15:03




$begingroup$
Why is $f^{-1}(V)={emptyset,X}$?
$endgroup$
– SmileyCraft
Jan 16 at 15:03




2




2




$begingroup$
@SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
$endgroup$
– drhab
Jan 16 at 15:05






$begingroup$
@SmileyCraft (typo, repaired) Because $V=varnothing$ or $V=Y={y}$ and $f:Xto Y$ can only be the constant function.
$endgroup$
– drhab
Jan 16 at 15:05














$begingroup$
Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
$endgroup$
– Uchiha
Jan 16 at 15:08




$begingroup$
Right. I did not consider this degenerate situation. Can one say anything about the case where $Y$ has more than one element?
$endgroup$
– Uchiha
Jan 16 at 15:08












$begingroup$
Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
$endgroup$
– drhab
Jan 16 at 15:14






$begingroup$
Here is another degenerate situation: let $Y$ have indiscrete topology. Then again $f^{-1}(V)in{varnothing,X}$ for every open $V$, and $Y$ can have as much elements as you want.
$endgroup$
– drhab
Jan 16 at 15:14














$begingroup$
Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
$endgroup$
– Uchiha
Jan 16 at 15:25






$begingroup$
Right, it seems the question is only interesting if more regularity is assumed on the topology of Y.
$endgroup$
– Uchiha
Jan 16 at 15:25




















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