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I'm only interested in the case where $A$ is a single point and $X=S^2$ for the following question from Hatcher's book on Algebraic Topology.

Question: Compute the homology of groups $H_n(X,A)$ when $X$ is $S^2$ or $T^2$ and $A$ is a finite set of points in $X$.

Answer Attempt: In Hatcher p.g. 117 it says you can think of the relative homology $H_n(X,A)$ as the measure of the difference between the two groups $H_n(X)$ and $H_n(A)$. Taking that literally, let $x\in X$, then $H_n(S^2\setminus {x})\cong H_n(\mathbb{R^2})\cong 0$

EDIT: Actually, if I were to take the statement, "you can think of the relative homology $H_n(X,A)$ as the measure of the difference between the two groups $H_n(X)$ and $H_n(A)$.", from Hatcher literally it would be more correct to note that since $H_2(X)\cong \mathbb{Z}$ and $H_2(A)=H_1(A)=0$ their difference is $\mathbb{Z}$.

Bob
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  • The equation in your attempt doesn't follow at all. The group $H_n(X, A)$ is not the difference or quotient of $H_n(X)$ and $H_n(A)$; they fit into a long exact sequence. Homology does not commute with taking quotients; the relative groups $H_n(X, A)$ indicate that failure. – anomaly Apr 06 '16 at 06:23
  • OK, so I should just stick with the conventional approach, namely, $0\to\mathbb{Z}\to H_2(X,A)\to 0 \implies H_2(X,A)=\mathbb{Z}$ – Bob Apr 06 '16 at 06:35
  • Right. Hatcher's remark is just a fuzzy statement for intuition; it shouldn't be taken literally. (For one thing, the situation gets more complicated with torsion.) – anomaly Apr 06 '16 at 15:26
  • one can think $S^2/A=S^2\bigvee\vee _{(|A|-1)}S^1$. – Biplab Feb 20 '24 at 03:39

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