Please, forgive me if I say something silly. I would like to prove that for any smooth compact $n$-dimensional sub-manifold $M$ of the open $n$-ball $B$ there is a finite $(n-1)$-dimensional simplicial complex $C\subset M$ such that $id_{M}$ is homotopic in $M$ to some map $f:M\to C$. Ideally, $C$ would be a (weak)-deformation retract of $M$.
It feels to me that in fact one should be able to show that every compact triangulable $n$-dimensional manifold $M$ all of whose components have non-trivial boundary deformation-retracts onto an $(n-1)$-dimensional simplicial complex, which would apply to the situation above, since any smooth manifold is triangulable.
The argument would be as follows. We construct a sequence of simplicial complexes $(T_{i})_{i\geq 0}$ starting with $T_{0}=T$. At step $i$ we choose, if available, some $(n-1)$-simplex $\sigma$ of $T_{i}$ that is contained in exactly one $n$-simplex $\tau$ of $T_{i}$. There is an obvious deformation of $|\tau|$ onto $|\partial\tau|\setminus|\sigma|$, which extends by the identity to a deformation of $|T_{i}|$ onto $|T_{i+1}|$, where $T_{i+1}=T_{i}\setminus\{\tau,\sigma\}$. If such $\sigma$ does not exist, then we stop. One simply observes that when the process stops there cannot be any $n$-simplices left, since otherwise the sum of all of them would constitute a non-trivial singular $n$-cylce with $\mathbb{Z}/2\mathbb{Z}$-coefficients in $|T_{i}|$ and thus in $M$, contradicting the fact that any component of $M$ has non-empty boundary.
I have a bad feeling about this, for some reason. Is there any problem with the previous paragraph? Or perhaps a more direct argument one could apply? Could one remove the triangulability assumption at the cost of only being able to require $C$ to be $(n-1)$-dimensional CW-subcomplex of $M$?