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!

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    $\begingroup$ Why are the fundamental groups Abelian? $\endgroup$ 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.) $\endgroup$ 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. $\endgroup$ Mar 3 '15 at 20:40
  • $\begingroup$ @AlexDegtyarev: Thanks for reminder. Here $C\geq 1$ is an absolute constant. The interesting case is actually $C>1$. $\endgroup$ 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. $\endgroup$ Mar 3 '15 at 23:38

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.


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