Mixing
On topological space whose homology groups/modules are trivial
$begingroup$
Let $X$ be a simply connected topological space. Consider the following three statements :
(1) $X$ is contractible.
(2) For every commutative ring $R$, $H_0(X,R)=R$ and $H_n(X,R)=0, forall n ge 1$
(3) $H_0(X,mathbb Z)=mathbb Z$ and $H_n(X,mathbb Z)=0,forall n ge 1$
Now we know that $(1) implies (2) implies (3)$.
My question is: Is any of these implications reversible ?
algebraic-topology homology-cohomology homotopy-theory
asked Jan 10 at 7:21
user521337user521337
1,1451416
$endgroup$
add a comment |
$begingroup$
Let $X$ be a simply connected topological space. Consider the following three statements :
(1) $X$ is contractible.
(2) For every commutative ring $R$, $H_0(X,R)=R$ and $H_n(X,R)=0, forall n ge 1$
(3) $H_0(X,mathbb Z)=mathbb Z$ and $H_n(X,mathbb Z)=0,forall n ge 1$
Now we know that $(1) implies (2) implies (3)$.
My question is: Is any of these implications reversible ?
algebraic-topology homology-cohomology homotopy-theory
asked Jan 10 at 7:21
user521337user521337
1,1451416
$endgroup$
add a comment |
$begingroup$
Let $X$ be a simply connected topological space. Consider the following three statements :
(1) $X$ is contractible.
(2) For every commutative ring $R$, $H_0(X,R)=R$ and $H_n(X,R)=0, forall n ge 1$
(3) $H_0(X,mathbb Z)=mathbb Z$ and $H_n(X,mathbb Z)=0,forall n ge 1$
Now we know that $(1) implies (2) implies (3)$.
My question is: Is any of these implications reversible ?
algebraic-topology homology-cohomology homotopy-theory
asked Jan 10 at 7:21
user521337user521337
1,1451416
$endgroup$
Let $X$ be a simply connected topological space. Consider the following three statements :
(1) $X$ is contractible.
(2) For every commutative ring $R$, $H_0(X,R)=R$ and $H_n(X,R)=0, forall n ge 1$
(3) $H_0(X,mathbb Z)=mathbb Z$ and $H_n(X,mathbb Z)=0,forall n ge 1$
Now we know that $(1) implies (2) implies (3)$.
My question is: Is any of these implications reversible ?
algebraic-topology homology-cohomology homotopy-theory
algebraic-topology homology-cohomology homotopy-theory
asked Jan 10 at 7:21
user521337user521337
1,1451416
asked Jan 10 at 7:21
user521337user521337
1,1451416
asked Jan 10 at 7:21
user521337user521337
1,1451416
asked Jan 10 at 7:21
user521337user521337
1,1451416
asked Jan 10 at 7:21
user521337user521337
1,1451416
1,1451416
add a comment |
add a comment |
2 Answers
2
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By the Hurewicz theorem, (3) (together with simple connectedness) implies all homotopy groups of $X$ are trivial (if some homotopy group is nontrivial, the least nontrivial homotopy group would give a nontrivial homology group). If $X$ has the homotopy type of a CW-complex, then $X$ is contractible by Whitehead's theorem. So both implications are reversible if $X$ has the homotopy type of a CW-complex.
For arbitrary simply connected spaces, the second implication is reversible but not the first. Indeed, (3) implies (2) immediately by the universal coefficient theorem. Some examples of simply connected spaces that satisfy (3) (and thus (2)) but are not contractible are the (open) long line and the Warsaw circle.
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
add a comment |
$begingroup$
(3) implies (2) by the
Universal Coefficient Theorem.
I'm pretty sure that (2) does not imply (1); that would make topology too easy....
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the Hurewicz theorem, (3) (together with simple connectedness) implies all homotopy groups of $X$ are trivial (if some homotopy group is nontrivial, the least nontrivial homotopy group would give a nontrivial homology group). If $X$ has the homotopy type of a CW-complex, then $X$ is contractible by Whitehead's theorem. So both implications are reversible if $X$ has the homotopy type of a CW-complex.
For arbitrary simply connected spaces, the second implication is reversible but not the first. Indeed, (3) implies (2) immediately by the universal coefficient theorem. Some examples of simply connected spaces that satisfy (3) (and thus (2)) but are not contractible are the (open) long line and the Warsaw circle.
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
add a comment |
$begingroup$
By the Hurewicz theorem, (3) (together with simple connectedness) implies all homotopy groups of $X$ are trivial (if some homotopy group is nontrivial, the least nontrivial homotopy group would give a nontrivial homology group). If $X$ has the homotopy type of a CW-complex, then $X$ is contractible by Whitehead's theorem. So both implications are reversible if $X$ has the homotopy type of a CW-complex.
For arbitrary simply connected spaces, the second implication is reversible but not the first. Indeed, (3) implies (2) immediately by the universal coefficient theorem. Some examples of simply connected spaces that satisfy (3) (and thus (2)) but are not contractible are the (open) long line and the Warsaw circle.
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
$endgroup$
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
add a comment |
$begingroup$
By the Hurewicz theorem, (3) (together with simple connectedness) implies all homotopy groups of $X$ are trivial (if some homotopy group is nontrivial, the least nontrivial homotopy group would give a nontrivial homology group). If $X$ has the homotopy type of a CW-complex, then $X$ is contractible by Whitehead's theorem. So both implications are reversible if $X$ has the homotopy type of a CW-complex.
For arbitrary simply connected spaces, the second implication is reversible but not the first. Indeed, (3) implies (2) immediately by the universal coefficient theorem. Some examples of simply connected spaces that satisfy (3) (and thus (2)) but are not contractible are the (open) long line and the Warsaw circle.
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
$endgroup$
By the Hurewicz theorem, (3) (together with simple connectedness) implies all homotopy groups of $X$ are trivial (if some homotopy group is nontrivial, the least nontrivial homotopy group would give a nontrivial homology group). If $X$ has the homotopy type of a CW-complex, then $X$ is contractible by Whitehead's theorem. So both implications are reversible if $X$ has the homotopy type of a CW-complex.
For arbitrary simply connected spaces, the second implication is reversible but not the first. Indeed, (3) implies (2) immediately by the universal coefficient theorem. Some examples of simply connected spaces that satisfy (3) (and thus (2)) but are not contractible are the (open) long line and the Warsaw circle.
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
edited Jan 10 at 7:44
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
answered Jan 10 at 7:30
Eric WofseyEric Wofsey
185k13213339
185k13213339
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
add a comment |
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
just out of curiosity ; do you have a metrizable counterexample ?
$endgroup$
– user521337
Jan 10 at 8:13
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
$begingroup$
The Warsaw circle is metrizable; it's a subspace of $mathbb{R}^2$.
$endgroup$
– Eric Wofsey
Jan 10 at 16:07
add a comment |
$begingroup$
(3) implies (2) by the
Universal Coefficient Theorem.
I'm pretty sure that (2) does not imply (1); that would make topology too easy....
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
(3) implies (2) by the
Universal Coefficient Theorem.
I'm pretty sure that (2) does not imply (1); that would make topology too easy....
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
add a comment |
$begingroup$
(3) implies (2) by the
Universal Coefficient Theorem.
I'm pretty sure that (2) does not imply (1); that would make topology too easy....
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
(3) implies (2) by the
Universal Coefficient Theorem.
I'm pretty sure that (2) does not imply (1); that would make topology too easy....
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 10 at 7:28
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
add a comment |
add a comment |
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