Consider two compact, oriented and connected manifolds $\mathcal{M},\mathcal{N}$ with possibly non-empty connected boundaries $\partial\mathcal{M}$ and $\partial\mathcal{N}$. Now, in some project, I encounted the following manifold:

$$\mathcal{Q}:=(\mathcal{M}\# B^{d})\#_{\partial}\mathcal{N}$$

Let me briefly explain the notation I used for defining the manifold $\mathcal{Q}$:

  • $B^{d}$ denotes the closed $d$-dimensional ball, whose boundary is the $(d-1)$-sphere $S^{d-1}$.
  • $\#$ denotes the internal, oriented connected sum, i.e. the manifold obtained by cutting out two internal $d$-balls not touching the boundaries of two manifolds with connected boundary and gluing the created boundary spheres together via an orientation-reversing homeomorphism.
  • $\#_{\partial}$ denotes the oriented boundary-connected sum, i.e. the manifold obtained by cutting out two $(d-1)$-balls living purely on the boundaries of two manifolds with connected boundary and gluing them together via an orientation-reversing homeomorphism.

Now to my question:

In the special case where $\mathcal{M}$ has empty boundary, i.e. $\partial\mathcal{M}=\emptyset$, is it true that $\mathcal{Q}\cong \mathcal{M}\#\mathcal{N}$?

Of course, in the trivial case where $\mathcal{M}$ is homeomorphic to the $d$-sphere $S^{d}$, this is trivially true, since

$$\mathcal{Q}=(S^{d}\# B^{d})\#_{\partial}\mathcal{N}\cong B^{d}\#_{\partial}\mathcal{N}\cong \mathcal{N}\cong S^{d}\#\mathcal{N}.$$

When I think about some very simple (but non-trivial) examples in low-dimensions, then I think it seems to be the case more generally. However, I have a lot of struggle imagining these things.

(Non-trivial example where it works: $\mathcal{M}=T^{2}$ (2-torus), $\mathcal{N}=S^{1}\times [0,1]$ (cylinder), then we get for $\mathcal{Q}$ as well as $\mathcal{M}\#\mathcal{N}$ the unique (up to homeomorphism) surface with genus=1 and number of boundary components=2)

What is clear is that the boundaries of $\mathcal{Q}$ and $\mathcal{M}\#\mathcal{N}$ are the same if $\mathcal{M}$ is closed, since then $\mathcal{M}\# B^{d}$ has boundary $S^{d-1}$, from which follows that the boundary of $\mathcal{Q}$ coincides with the boundary of $\mathcal{N}$. This is of course also the case for $\mathcal{M}\#\mathcal{N}$ and hence $$\partial\mathcal{Q}\cong\partial(\mathcal{M}\#\mathcal{N})\cong\partial\mathcal{N}.$$ However, this alone does of course not ensure that $\mathcal{Q}$ is homeomorphic to $\mathcal{M}\#\mathcal{N}$.

Remark: What I meant with "associativity" in the title is that, if my question turns out to be true, then we can write $$\mathcal{Q}=(\mathcal{M}\# B^{d})\#_{\partial}\mathcal{N}\cong \mathcal{M}\# (B^{d}\#_{\partial}\mathcal{N})\cong\mathcal{M}\#\mathcal{N}$$ whenever $\mathcal{M}$ is closed. So, this looks like some kind of associativity, although we used two different products.


1 Answer 1


This is true in the piecewise linear category. As you note, the boundary connect sum of $B$ and $N$ is homeomorphic to $N$. Now apply a result of Gugenheim [1953]: if $C$ and $D$ are $n$-balls embedded in the interior of a manifold, then there is an isotopy taking $C$ to $D$. This obtains the middle homeomorphism in your last displayed equation.

  • $\begingroup$ Thank you very much! Could you elaborate a little bit more on how exactly this shows the claim? Unfortunately, I am not able to find the cited paper bby Gugenheim... $\endgroup$
    – B.Hueber
    Commented Jan 6, 2022 at 13:19
  • $\begingroup$ See Theorem 3 of the first paper here: londmathsoc.onlinelibrary.wiley.com/action/… $\endgroup$
    – Sam Nead
    Commented Jan 6, 2022 at 21:32

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