Quotient map between homotopy equivalent spaces
$begingroup$
$require{AMScd}$
Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
begin{CD}
X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
Is it true that $q$ is then a homotopy equivalence?
Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?
general-topology
asked Dec 17 '18 at 16:47
BertrandBertrand
685
$endgroup$
add a comment |
$begingroup$
$require{AMScd}$
Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
begin{CD}
X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
Is it true that $q$ is then a homotopy equivalence?
Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?
general-topology
asked Dec 17 '18 at 16:47
BertrandBertrand
685
$endgroup$
add a comment |
$begingroup$
$require{AMScd}$
Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
begin{CD}
X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
Is it true that $q$ is then a homotopy equivalence?
Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?
general-topology
asked Dec 17 '18 at 16:47
BertrandBertrand
685
$endgroup$
$require{AMScd}$
Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
begin{CD}
X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
Is it true that $q$ is then a homotopy equivalence?
Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?
general-topology
general-topology
asked Dec 17 '18 at 16:47
BertrandBertrand
685
asked Dec 17 '18 at 16:47
BertrandBertrand
685
asked Dec 17 '18 at 16:47
BertrandBertrand
685
asked Dec 17 '18 at 16:47
BertrandBertrand
685
asked Dec 17 '18 at 16:47
BertrandBertrand
685
685
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
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add a comment |
$begingroup$
Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:
If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
add a comment |
$begingroup$
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
add a comment |
$begingroup$
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
$endgroup$
There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).
Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.
Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
edited Dec 17 '18 at 23:11
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
answered Dec 17 '18 at 22:21
Pavel ČoupekPavel Čoupek
4,45611126
4,45611126
add a comment |
add a comment |
$begingroup$
Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:
If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
$endgroup$
add a comment |
$begingroup$
Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:
If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
$endgroup$
add a comment |
$begingroup$
Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:
If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
$endgroup$
Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:
If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
answered Dec 17 '18 at 17:00
positrón0802positrón0802
4,353520
4,353520
add a comment |
add a comment |
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