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Fix $n \geq 1$ and let $B$ denote a triangulated closed $n$-ball. Let $D$ be a subset of the boundary of $B^n$ that is homeomorphic to the closed $(n-1)$-ball and such that is properly triangulated by the same triangulation as well.

I would like to compute the Betti numbers of $B$ relative to $D$. I have tried to regard the long exact sequence on homology of the chain complexes, and that the $0$-th homology is trivial as it is isomorphic to the $0$-th homology group of the de-Rham complex. Is it correct that all Betti number of $B$ relative to $D$ vanish?

I am not used to algebraic topology, and need the result only in a secondary (if even that) context. I could use a reference which clarifies under which circumstances to use the long exact sequence on homology in the context of simplicial complexes.

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Yes, you can use the long exact sequence of relative homology. In fact, both the ball and the disk are contractible so they only have nonzero homology in degree 0, and there the inclusion is an isomorphism on homology. Examining the long exact sequence, this is enough to conclude that the relative homology is always zero. (So the Betti numbers all vanish.)

You can also see this by noting that there is a homotopy equivalence between the pair (Ball,disk) and (Disk, Disk) essentially by crushing the non-disk parts of the ball down onto the boundary disk. Then your statement holds by the homotopy invariance of homology.

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