Example of $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$












1












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In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.



Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?






algebraic-topology continuity







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  • 1




    $begingroup$
    Think first about maps of intervals constructed via Peano curves.
    $endgroup$
    – Moishe Kohan
    Dec 24 '18 at 21:02
















1












$begingroup$


In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.



Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?






algebraic-topology continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Think first about maps of intervals constructed via Peano curves.
    $endgroup$
    – Moishe Kohan
    Dec 24 '18 at 21:02














1












1








1





$begingroup$


In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.



Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?






algebraic-topology continuity







share|cite|improve this question











$endgroup$




In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.



Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?






algebraic-topology continuity




algebraic-topology continuity






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share|cite|improve this question













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edited Dec 25 '18 at 7:24









Saad

21k92453




21k92453










asked Dec 24 '18 at 20:53









JaviJavi

3,29321133




3,29321133








  • 1




    $begingroup$
    Think first about maps of intervals constructed via Peano curves.
    $endgroup$
    – Moishe Kohan
    Dec 24 '18 at 21:02














  • 1




    $begingroup$
    Think first about maps of intervals constructed via Peano curves.
    $endgroup$
    – Moishe Kohan
    Dec 24 '18 at 21:02








1




1




$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02




$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02










1 Answer
1






active

oldest

votes


















5












$begingroup$

Here's an example for $n=1$.



Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.



One can use similar tricks for any $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It wasn't as hard as I thought! Thanks
    $endgroup$
    – Javi
    Dec 24 '18 at 21:08












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Here's an example for $n=1$.



Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.



One can use similar tricks for any $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It wasn't as hard as I thought! Thanks
    $endgroup$
    – Javi
    Dec 24 '18 at 21:08
















5












$begingroup$

Here's an example for $n=1$.



Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.



One can use similar tricks for any $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It wasn't as hard as I thought! Thanks
    $endgroup$
    – Javi
    Dec 24 '18 at 21:08














5












5








5





$begingroup$

Here's an example for $n=1$.



Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.



One can use similar tricks for any $n$.






share|cite|improve this answer









$endgroup$



Here's an example for $n=1$.



Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.



One can use similar tricks for any $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 21:05









Lord Shark the UnknownLord Shark the Unknown

109k1163136




109k1163136












  • $begingroup$
    It wasn't as hard as I thought! Thanks
    $endgroup$
    – Javi
    Dec 24 '18 at 21:08


















  • $begingroup$
    It wasn't as hard as I thought! Thanks
    $endgroup$
    – Javi
    Dec 24 '18 at 21:08
















$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08




$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08


















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