Skip to main content
added 71 characters in body; added 46 characters in body
Source Link
Eric Wofsey
  • 31.2k
  • 2
  • 115
  • 151

Let $X$ be any finite complex such that any map $X\to X$ homotopic to the identity is surjective and for which there is a surjective but nullhomotopic PL map $f:X\to X$ (for instance, and let$X$ could be $S^n$ for any $n>0$). Let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself. Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.

For $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Let $X$ be any finite complex for which there is a surjective but nullhomotopic PL map $f:X\to X$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself. Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.

For $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Let $X$ be any finite complex such that any map $X\to X$ homotopic to the identity is surjective and for which there is a surjective but nullhomotopic PL map $f:X\to X$ (for instance, $X$ could be $S^n$ for any $n>0$). Let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself. Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.

For $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

added 50 characters in body
Source Link
Eric Wofsey
  • 31.2k
  • 2
  • 115
  • 151

TakeLet $X$ be any finite complex for which there is a surjective but nullhomotopic PL map $f:S^1\to S^1$$f:X\to X$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $S^1$$X$ is $K$ itself. Indeed, in order to be able to contract the first circlecopy of $X$, you must take a subcomplex that contains the entire second circlecopy, and to contract the second circlecopy, you must contain the third circlecopy, and so on.

YouFor $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Take a surjective but nullhomotopic PL map $f:S^1\to S^1$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $S^1$ is $K$ itself. Indeed, in order to be able to contract the first circle, you must take a subcomplex that contains the entire second circle, and to contract the second circle, you must contain the third circle, and so on.

You can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Let $X$ be any finite complex for which there is a surjective but nullhomotopic PL map $f:X\to X$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself. Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.

For $X=S^1$, you can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

added 439 characters in body
Source Link
Eric Wofsey
  • 31.2k
  • 2
  • 115
  • 151

Take a surjective but nullhomotopic PL map $f:S^1\to S^1$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $S^1$ is $K$ itself. Indeed, in order to be able to contract the first circle, you must take a subcomplex that contains the entire second circle, and to contract the second circle, you must contain the third circle, and so on.

You can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Take a surjective but nullhomotopic PL map $f:S^1\to S^1$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $S^1$ is $K$ itself. Indeed, in order to be able to contract the first circle, you must take a subcomplex that contains the entire second circle, and to contract the second circle, you must contain the third circle, and so on.

Take a surjective but nullhomotopic PL map $f:S^1\to S^1$, and let $K$ be the mapping telescope obtained by iterating $f$. Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $S^1$ is $K$ itself. Indeed, in order to be able to contract the first circle, you must take a subcomplex that contains the entire second circle, and to contract the second circle, you must contain the third circle, and so on.

You can even visualize this example pretty easily. Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C". Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly. Repeating this infinitely gives the telescope $K$.

Source Link
Eric Wofsey
  • 31.2k
  • 2
  • 115
  • 151
Loading