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.



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


