10
$\begingroup$

Whitehead's conjecture states that if $L$ is an aspherical 2-complex and $K$ is a subcomplex of $L$, then $K$ is also aspherical. It is known by work of Howie and Luft that if the Whitehead conjecture is false, then there is a counterexample where $L$ is constructed as the limit of a series of nested spaces $K= K_0 \subset K_1 \subset K_2 \subset \cdots$ where the inclusion of $K_i$ in $K_{i+1}$ is nullhomotopic and each $K_i$ is not aspherical (or should I just say spherical?).

I do not think about 2-complexes very often so I apologize if this is too basic, but I can't picture any examples of such inclusions of 2-complexes $A \subset B$ where the inclusion is null homotopic and $\pi_2(A) \neq 0$. What are some examples of this? Ideally, I would like for $A$ and $B$ to both be finite. Any example will probably have $\pi_2(B) \neq 0$, since otherwise you solve Whitehead.

Given a 2-complex $A$, can we always find another 2-complex $B$ containing $A$ as a subcomplex so that $A$ is nullhomotopic in $B$? If you could, then this would yield a solution to the Whitehead conjecture, since you could just do that over and over again, and the resulting limit space would be aspherical by compactness. So there are presumably obstructions, does anyone know some?

Edit: Whoops - In light of Mike's observation I should assume that $H_2(A) = 0$. For the setup of the Whitehead conjecture, we have $H_2(K) = 0$, so I definitely should have included that initially - sorry.

$\endgroup$
  • 5
    $\begingroup$ For the final question: if $A \subset B$ is an inclusion of 2-complexes, then $H_2(A)$ injects into $H_2(B)$ (as $H_3(A,B) = 0$ for dimensions reasons). So for instance $S^2$ can't be null as a subcomplex of any 2-complex. $\endgroup$ – Mike Miller Mar 27 at 2:43
  • 1
    $\begingroup$ I'm not certain that this question is any easier than the Whitehead conjecture itself! $\endgroup$ – HJRW Mar 31 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.