Action of the symmetric group on connected sums of manifolds (minus a disk)

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?

• 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. – Ryan Budney Mar 14 at 15:20
• @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? – Bashar Saleh Mar 14 at 16:13
• 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. – Ryan Budney Mar 14 at 16:53