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.










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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.










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      up vote
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      up vote
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      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.










      share|cite|improve this question













      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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      asked 2 days ago









      Stephen

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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.






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          • 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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          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.






          share|cite|improve this answer























          • 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















          up vote
          1
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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.






          share|cite|improve this answer























          • 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













          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














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          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


















          • 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


















           

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