Let $M$ be a connected compact topological $n$-dimensional manifold without a boundary and with a CW-structure $M= \bigcup M^i$. We have that $$ (\#^g M)\smallsetminus D^n \simeq \bigvee_{i=1}^gM^{n-1}$$ The symmetric group $S_g$ acts on $\bigvee_{i=1}^gM^{n-1}$ by permuting the summands. In particular every element $\sigma\in S_g$ induces an automorphism $\sigma_*\colon \bigvee_{i=1}^gM^{n-1}\to\bigvee_{i=1}^gM^{n-1}$, which extends (non-uniquely) to a homotopy self-equivalence of $(\#^g M)\smallsetminus D^n$.

My question: Could we extend $\sigma_*$ to be a homotopy self-equivalence of $(\#^g M)\smallsetminus D^n$ that preserves the boundary pointwise?

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    $\begingroup$ Yes, you can do this, and a little more. Think of your connect sum $\#^g M$ as having this construction: drill $g$ disjoint open balls out from $S^n$ and glue in $g$ copies of $M \setminus int(D^n)$. Notice that "the space of g disjoint open balls in $S^n$" is connected, and you can interchange balls. That gives you your lift $\sigma_*$. It won't generally be unique, but it exists. $\endgroup$ – Ryan Budney Mar 14 at 15:20
  • $\begingroup$ @RyanBudney, thank you! If i understood you correct, then, if I drill $g$ holes in $S^n$ there exists a (homotopy) self-equivalence of that space (i.e $S^n$ with $g$ holes) that permutes two holes but fixes the other holes pointwise? Is that something obvious? $\endgroup$ – Bashar Saleh Mar 14 at 16:13
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    $\begingroup$ Yes -- with a little extra information. Think about the "space of discs" that I describe as embeddings of discs into the sphere. Apply the isotopy-extension theorem to a 1-parameter family of such embeddings. $\endgroup$ – Ryan Budney Mar 14 at 16:53

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