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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