Homology Groups of $S^n$

Is it possible to consider $$S^n$$ as a $$0$$-simplex and a singular map of an $$n$$-simplex so that the $$n$$-simplex forms the surface of $$S^n$$ minus a point, and the point is the singular map of the $$0$$-simplex? Can a (singular map of an) $$n$$-simplex have zero boundary for $$n>1$$? I am trying to understand the homology groups of $$\mathbb{R}^n$$ with finitely many points removed. I imagine I could use the Mayer-Vietoris exact sequence and homotopy invariance, but I want to try to find the groups directly from the definition to confirm that I am right. So, I have started by showing that $$X=\mathbb{R}^n\backslash\{x_1,\ldots,x_m\}$$ is homotopy equivalent (deformation retracts) to the wedge sum of $$m$$ copies of $$S^{n-1}$$ meeting at a common point, as in the diagram for $$n=3,m=4$$.

• You say you want to do this "directly from the definition" but you do not say which definition. If you mean the definition of singular homology, it is most certainly not an easy exercise to compute these singular homology groups directly from the definition. Commented Feb 2, 2021 at 15:09
• I see, I suppose I am doing it by simplicial homology, which as far as I know is the same as singular homology, but I guess I haven't seen a proof of that yet. Commented Feb 2, 2021 at 15:17
• They are not "the same". They are isomorphic, but that is a major theorem. Commented Feb 2, 2021 at 16:42
• Now that I think of it, I think the only space whose homology I've ever seen directly computed completely using singular homology might be a point. Or maybe "two points". :) Commented Feb 3, 2021 at 0:48

Certainly you can construct $$S^n$$ as a cell complex with one 0-cell and one $$n$$-cell, and compute cellular homology that way. Many algebraic topology books use this as their first example of computing cellular homology.