Example of $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$
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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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add a comment |
$begingroup$
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
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1
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Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $n$.
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $n$.
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
$begingroup$
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 $n$.
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
$begingroup$
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 $n$.
$endgroup$
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 $n$.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02