I am trying to answer the following question:
$(a)$ Compute the homology groups $H_n(X, A)$ when $X$ is $S^2$ or $S^1 \times S^1$ and $A$ is a finite set of points in $X.$
And I am trying to use the following proposition in my answer:
But I do not know what will I get if I moded the sphere $S^2$ with $A$ and similarly for the torus, could anyone clarify this to me please?
Note that I tried to understand the method in the solution here Compute the homology groups $H_n(X, A)$ when $X$ is $\mathbb{S}^2$ or $\mathbb{S}^1\times \mathbb{S}^1$ and $A$ is a finite set of points in $X$. but I did not get it at all, so any explanation of it will also be helpful.
Another question about the linked solution, are they using the ideas in problem #16, the one just before #17 in AT
I'll assume that $A=\{a,b\}$ for simplicity. To extend this, you can really use induction because the argument will not change.
Here are two ways to get started. The first is to first recognize $S^2/A$ as homotopy equivalent to something familiar. The second is to leverage the long exact sequence of pairs.
Hint 1: In general if you have a space $X$ (say, a CW-complex) and you take a quotient by two points, this is homotopy equivalent to taking $X$ and pasting in a copy of $[0,1]$ attaching those two points. Said differently, $X/\{a,b\} \simeq (X \coprod I)/\{a \sim 0, b \sim 1\}$. So, this space is homotopy equivalent to $S^2$ with a noodle attaching two points. You can "slurp spaghetti" to show that this is in fact wedge sum of two familiar spaces. By slurping spaghetti, I mean use path connectedness to take a path from $a$ to $b$ in $S^2$ and contract along it.
Hint 2: $H_n(X,A)=\tilde{H}_n(X/A)$, and we have a long exact sequence $H_n(A) \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to \cdots$.
Computing $H_n(A)$ should be fine. For $H_n(S^2)$, you can use the suspension isomorphism $\tilde{H}_i(S^1) \cong \tilde{H}_{i+1}(S^2)$, Mayer Vietoris on $S^2=D^2 \cup D^2$, or explicitly try to use a $CW$ decomposition on $S^2$ and compute cellular homology.