# Kind of "associativity" of certain connected sum involving both manifolds with and without boundary

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.

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.