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I can prove easily in the case of $n$ is over zero but failed to prove when $n$ equals 0. I tried like below.

Since $(X_a, x_a)$ is a good pair, a pair of disjoint union of $X_a$ and disjoint union of $x_a$ is also good. So one can say that relative homology group regarding to pair $(D(X_a), D(x_a))$ is isomorphic to reduced homology group regarding to $D(X_a)/D(x_a)$. Here, $D(x_a)$ means disjoint union of $x_a$ and $D(X_a)$ has similar meaning. So it is clear that one should show that $H_n(D(X_a), D(x_a))$ is isomorphic to free abelian group which is indicated. When $n$ is over 0, one can show this by setting a long exact sequence of homology groups. But I think this method is impossible for the case $n=0$. Someone can help me?

user301120
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1 Answers1

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I'm not sure what you're trying to do with the long exact sequence, but I think this is the idea:

The quotient map $$\left(\coprod X_\alpha, \coprod \{x_\alpha\}\right) \to \left(\coprod X_\alpha / \coprod \{x_\alpha\}, \coprod \{x_\alpha\}/\coprod \{x_\alpha\}\right)$$ induces isomorphisms $$H_n\left(\coprod X_\alpha, \coprod \{x_\alpha\}\right) \cong \tilde H_n\left(\coprod X_\alpha / \coprod \{x_\alpha\}, \coprod \{x_\alpha\}/\coprod \{x_\alpha\}\right).$$ The right side is just $\tilde H_n\left(\bigvee X_\alpha\right)$ and the left side splits as $\bigoplus_\alpha H_n(X_\alpha, x_\alpha) \cong \bigoplus_\alpha \tilde H_n(X_\alpha)$.

manthanomen
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    Thank you for your answer. I failed to prove that the left one is isomorphic to $\bigoplus_\alpha H_n(X_\alpha, x_\alpha)$. Now, I succeed. – user301120 Dec 27 '15 at 10:12
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    How did you prove that? – Twnk Nov 11 '19 at 20:05
  • Disjoint union space is equipped with "disjoint union topology", which means each $X_\alpha$ is a path connected component (if each $X_\alpha$ is path connected) of $\coprod X_\alpha$. We can use the same procedure in Proposition 2.6 in Hatcher to prove the desired result. – HIGH QUALITY Male Human Being Dec 01 '22 at 22:54