TITLE: Connected sum of topological manifolds
QUESTION [43 upvotes]: A definition of the connected sum of two $n$-manifolds $M$ and $M'$  begins by considering two $n$-balls $B$ in $M$, $B'$ in $M'$, and glueing the varieties $M\setminus \mathring B$ and $M'\setminus \mathring B'$ along their boundary (an $(n-1)$-sphere) by an orientation-reversing homeomorphism.  The construction depends a priori on these various choices, but it is asserted at many places of the litterature (Lee's book on topological manifolds for example, as well as Wikipedia) 
 that the result does not depend on these choices.
In the differentiable case, a reference is given to a theorem of Palais (Natural operations on differential forms, Thm. 5.5) which asserts — roughly — that two embedding of $n$-balls differ by a global diffeomorphism which is isotopic to identity.
Are the details of this independence written somewhere in the litterature,
both in the continuous and the smooth case?

REPLY [12 votes]: The previous version of this post was incorrect.
EDIT:
I just taught a related fact in my undergraduate topology class (in the topological category), and in fact I usually give a version of this as a homework exercise (with hints). 
The idea is to define connected sum along the coordinate balls, i.e. the balls that are mapped via a fixed coordinate chart to standard balls in $\mathbb R^n$. For such balls the argument is easy, and no annulus theorem is needed because any two metric balls in $\mathbb R^n$ are ambiently isotopic by an isotopy that equals the identity outside a compact subset  (this can be done with bare hands). Since the homeomorphism group of manifold acts transitively on the manifold (nice exercise), any two points can be brought into the same chart. This provides some clue (not a complete proof) why the connected sum operation is independent of the choices involved.