The question is as in the title:

Is every finite subcomplex of a contractible simplicial complex $K$ contained in a finite contractible subcomplex of $K$? What if we are allowed to take subdivisions? What if the complex is locally compact and/or finitely dimensional?

I suppose the answer is no, but I could not find any counterexamples.

  • 1
    $\begingroup$ Take a simplicial model for $S^n$ contained in $S^{\infty}$. Does this not give a counterexample? $\endgroup$ Jan 21, 2013 at 23:40
  • 2
    $\begingroup$ Isn't it contained in the half-sphere of $S^{n+1}$? $\endgroup$ Jan 22, 2013 at 7:06

1 Answer 1


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 $K$.


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