4
$\begingroup$

Let \begin{equation} A\stackrel{\alpha}{\longrightarrow}B\stackrel{\beta}{\longrightarrow}C\quad\quad (1) \end{equation} be a short sequence of (abelian or nonabelian) groups and homomorphisms. We say that the sequence is fast if $\ker(\beta\alpha)=\ker \alpha$ or, equivalently, $\ker(\beta\alpha)\subset \ker \alpha$. Dually, the sequence is slow if ${\rm im}(\beta\alpha)={\rm im}\,\beta$ or, equivalently, ${\rm im}\,\beta\subset {\rm im}(\beta\alpha)$. In particular, the short sequence in (1) is fast if $\alpha=0$ and slow if $\beta=0$.

Now, let $\Omega\subset\mathbb{R}^3$ be a topological ball (or even stronger $\overline{\Omega}=f(\overline{\mathbb{B}})$, where $f$ is a homeomorphism from the closed unit ball $\overline{\mathbb{B}}$ onto $\overline{\Omega}$). Suppose that for any $x\in \mathbb{R}^3$, $r>0$, we know that the sequence ($C\geq 1$ is an absolute constant) \begin{align*} H_1(\Omega\cap B(a,r))\rightarrow H_1(\Omega\cap B(a,Cr))\rightarrow H_1(\Omega) \end{align*} is fast (the group homomorphism is induced by the inclusion). Is it true that the sequence \begin{align*} \pi_1(\Omega\cap B(a,r))\rightarrow \pi_1(\Omega\cap B(a,Cr))\rightarrow \pi_1(\Omega) \end{align*} is fast as well?

Similarly, if we know that the sequence \begin{align*} H_1(\Omega\backslash B(a,Cr))\rightarrow H_1(\Omega\backslash B(a,r))\rightarrow H_1(\Omega) \end{align*} is fast (the group homomorphism is induced by the inclusion). Is it true that the sequence \begin{align*} \pi_1(\Omega\backslash B(a,Cr))\rightarrow \pi_1(\Omega\backslash B(a,r))\rightarrow \pi_1(\Omega) \end{align*} is fast as well? Above, $H_1$ and $\pi_1$ are the first homology group and fundamental group, respectively.

Here, the assumption $n=3$ is very important since for $n\geq 4$, it is easy to cook up a counterexample (using the Artin-Fox arc or Poincaré sphere).

Comments, suggestions are warmly welcome!

$\endgroup$
  • 4
    $\begingroup$ Why are the fundamental groups Abelian? $\endgroup$ – Gabriel C. Drummond-Cole Mar 3 '15 at 7:43
  • $\begingroup$ In your settings, $C=0$ (a ball), so "fast" seems to mean $\alpha=0$. Then, it looks like you can get a counterexample to (1) using Alexender's horned sphere. (Though, here you may have a problem with extending the homeomorphism to the closure.) $\endgroup$ – Alex Degtyarev Mar 3 '15 at 9:05
  • $\begingroup$ @GabrielC.Drummond-Cole: I agree that there is no reason why the implication should hold, but for $n=3$, it seems that it is not easy to use the "wild" examples from topology to serve as a counter-example and so I would like to know whether this is some point hidden for this. $\endgroup$ – Changyu Guo Mar 3 '15 at 20:40
  • $\begingroup$ @AlexDegtyarev: Thanks for reminder. Here $C\geq 1$ is an absolute constant. The interesting case is actually $C>1$. $\endgroup$ – Changyu Guo Mar 3 '15 at 20:42
  • $\begingroup$ It's very easy to come up with an $\Omega$ which is not wild at all and whose intersection with a standard ball is a wedge of circles. $\endgroup$ – Gabriel C. Drummond-Cole Mar 3 '15 at 23:38
1
$\begingroup$

This is not an answer but you asked for an example and I couldn't give a picture in a comment.

I objected that the fundamental groups might not be Abelian and suggested that it was easy for the intersections to be homotopy equivalent to wedges of circles. Here is a picture, viewed in cutaway. cylinder with two hemispherical depressions in one end

Intersecting with a standard sphere directly below this shape would give something with non-Abelian fundamental group.

This is not supposed to be a counterexample to your question, just indication that your question is not well-posed because you only defined being fast or slow for Abelian groups. But let's say that I make up a definition of what you mean when $\pi_1$ is not Abelian. Then I'm confused because I think $\pi_1(\Omega)$ and $H_1(\Omega)$ are always $0$. Then it seems like you're asking whether, for the map $\Omega\cap B(a,Cr)\to \Omega\cap B(a,r)$ being zero on $H_1$ implies being zero on $\pi_1$. But that is a much simpler question, so I think I must be misunderstanding something.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.