A non-polyhedral pair $(X,A)$ with both $X$ and $A$ polyhedral
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Does there exist a topological space $X$ and a closed subspace $A$ such that each of $X$ and $A$ is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair $(S,T)$ consisting of a simplicial complex $S$ and a subcomplex $T$ with $(X,A) cong (|S|,|T|)$?
This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space $A$ inside the unit square $X$, but it seems that $A$ is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer.
general-topology algebraic-topology simplicial-complex
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Does there exist a topological space $X$ and a closed subspace $A$ such that each of $X$ and $A$ is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair $(S,T)$ consisting of a simplicial complex $S$ and a subcomplex $T$ with $(X,A) cong (|S|,|T|)$?
This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space $A$ inside the unit square $X$, but it seems that $A$ is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer.
general-topology algebraic-topology simplicial-complex
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Does there exist a topological space $X$ and a closed subspace $A$ such that each of $X$ and $A$ is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair $(S,T)$ consisting of a simplicial complex $S$ and a subcomplex $T$ with $(X,A) cong (|S|,|T|)$?
This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space $A$ inside the unit square $X$, but it seems that $A$ is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer.
general-topology algebraic-topology simplicial-complex
Does there exist a topological space $X$ and a closed subspace $A$ such that each of $X$ and $A$ is isomorphic to the topological space of some simplicial complex, but such that there does not exist a pair $(S,T)$ consisting of a simplicial complex $S$ and a subcomplex $T$ with $(X,A) cong (|S|,|T|)$?
This is (essentially) problem A2 from Chapter 3 of Spanier's book Algebraic topology. At first I suspected that the intended example was the comb space $A$ inside the unit square $X$, but it seems that $A$ is not polyhedral (if it were, then being compact, it would have to be the space of a finite simplicial complex). It doesn't seem to me that there is an example that has already appeared in Spanier's book, so I am somewhat at a loss as to the intended answer.
general-topology algebraic-topology simplicial-complex
general-topology algebraic-topology simplicial-complex
asked 2 days ago
Stephen
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10.4k12237
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One example is the Alexander horned sphere. This is a pathological embedding of $S^2$ into $mathbb{R}^3$, which in particular has the property that the unbounded component of its complement has non-finitely generated fundamental group. If you take $A$ to be the horned sphere and $X$ to be a closed ball containing it, then $A$ and $X$ are both finite simplicial complexes, but one of the components of $Xsetminus A$ has non-finitely generated fundamental group. If $X$ had a simplicial structure such that $A$ was a subcomplex, then this simplicial structure could only have finitely many cells since $X$ is compact, and then it is easy to see that $pi_1(Xsetminus A)$ would be finitely generated for any basepoint.
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One example is the Alexander horned sphere. This is a pathological embedding of $S^2$ into $mathbb{R}^3$, which in particular has the property that the unbounded component of its complement has non-finitely generated fundamental group. If you take $A$ to be the horned sphere and $X$ to be a closed ball containing it, then $A$ and $X$ are both finite simplicial complexes, but one of the components of $Xsetminus A$ has non-finitely generated fundamental group. If $X$ had a simplicial structure such that $A$ was a subcomplex, then this simplicial structure could only have finitely many cells since $X$ is compact, and then it is easy to see that $pi_1(Xsetminus A)$ would be finitely generated for any basepoint.
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
add a comment |
up vote
1
down vote
One example is the Alexander horned sphere. This is a pathological embedding of $S^2$ into $mathbb{R}^3$, which in particular has the property that the unbounded component of its complement has non-finitely generated fundamental group. If you take $A$ to be the horned sphere and $X$ to be a closed ball containing it, then $A$ and $X$ are both finite simplicial complexes, but one of the components of $Xsetminus A$ has non-finitely generated fundamental group. If $X$ had a simplicial structure such that $A$ was a subcomplex, then this simplicial structure could only have finitely many cells since $X$ is compact, and then it is easy to see that $pi_1(Xsetminus A)$ would be finitely generated for any basepoint.
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
One example is the Alexander horned sphere. This is a pathological embedding of $S^2$ into $mathbb{R}^3$, which in particular has the property that the unbounded component of its complement has non-finitely generated fundamental group. If you take $A$ to be the horned sphere and $X$ to be a closed ball containing it, then $A$ and $X$ are both finite simplicial complexes, but one of the components of $Xsetminus A$ has non-finitely generated fundamental group. If $X$ had a simplicial structure such that $A$ was a subcomplex, then this simplicial structure could only have finitely many cells since $X$ is compact, and then it is easy to see that $pi_1(Xsetminus A)$ would be finitely generated for any basepoint.
One example is the Alexander horned sphere. This is a pathological embedding of $S^2$ into $mathbb{R}^3$, which in particular has the property that the unbounded component of its complement has non-finitely generated fundamental group. If you take $A$ to be the horned sphere and $X$ to be a closed ball containing it, then $A$ and $X$ are both finite simplicial complexes, but one of the components of $Xsetminus A$ has non-finitely generated fundamental group. If $X$ had a simplicial structure such that $A$ was a subcomplex, then this simplicial structure could only have finitely many cells since $X$ is compact, and then it is easy to see that $pi_1(Xsetminus A)$ would be finitely generated for any basepoint.
edited 2 days ago
answered 2 days ago
Eric Wofsey
175k12202326
175k12202326
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
add a comment |
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
Thanks! (+1) Actually, I mentioned this example to my students, but I don't think it's quite fair. This can't possibly have been Spanier's intention, given the material covered so far...
– Stephen
2 days ago
add a comment |
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