Suppose we get two triangulations of a manifold with boundary $M$ such that the triangulation is compatible with boundary, i.e. the restriction on the boundary is itself a triangulation, is it these two singular simplicial complexes associated to the triangulations homologous? Or maybe we can ask the same problem for triangulable space.
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2$\begingroup$ What does it mean for a triangulation to be "compatible with the boundary"? I think your question is usually answered in textbooks that study Poincare duality from the cap product perspective. $\endgroup$– Ryan BudneyCommented Jul 6, 2022 at 6:23
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I'm not completely sure what you're asking, since there are a few things in your question that are unclear. For example, when one talks about chains being homologous, usually the interest is in cycles being homologous so that they represent the same elements of a homology group. I'm not sure what homologous would mean for non-cycles. If your manifold has boundary and you're thinking of the top dimensional chain given by the triangulation, then it won't be a cycle. However, if $M$ is compact and oriented, then the sum over all the $n=dim(M)$ simplices, suitably oriented, will represent the fundamental class $[M]\in H_n(M, \partial M)$. If you then choose a different triangulation but again take the sum over all n-simplices with the same compatible orientation, that will represent the same element of $H_n(M,\partial M)$.
If you want to think of these as singular chains, that can be done, but there are some technicalities involved in determining how a geometric simplex determines a singular simplex. You can read about that in most introductory book on homology theory. If you do that carefully, then yes the two resulting singular chains will be homologous modulo $\partial M$ since they represent the same relative singular homology class.
