Let $X=S^n, n>0$ and $A\subseteq X$ such that $X\setminus A$ contains finitely many points $x_1,...,x_m$. I need to show that the reduced homology $H_n(X,A)=\mathbb{Z}^m$.
I need to do this particularly by constructing a CW-complex on $X$, and not using anything fancy like excision. I tried many things and am utterly confused as to how to approach this.
If we take every point not in $A$ to be 0-cells, would $A$ be a CW-complex with no zero cells? How would I compute $H_i(A)$ as I don't know whether or not $A$ is a CW-complex (which prevents me form using the LES for relative homology). Hatcher doesn't have any helpful information on this problem.
In the case $n=1, m=2$, $X=S^1$ is a circle, and we take $x_1,x_2$ as the 0-cells, along with 2 arcs as 1-cells. Then $A$ is the disjoint union of 2 copies of $int D^1$ (ie. line segments). Is $A$ a CW-complex? Does $H_0(A)$ vanish in this case (since A has no 0-cells)? It shouldn't as then the LES would yield $H_1(X,A)=\mathbb{Z}$, and not $\mathbb{Z}^2$.
Can anyone please help me as to which CW complex on $X$ is needed? Appreciate it!