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I want to show that $\tilde{H_n}(\vee_{\alpha}X_{\alpha}) = \oplus \tilde{H_n}(X_{\alpha})$.

Note that $\tilde{H_n}(\vee_{\alpha}X_{\alpha}) = H_n(\vee_{\alpha}X_{\alpha}, x_0)$ where $x_0$ is the basepoint. Hatcher hinted that we should consider $H_n(\sqcup X_{\alpha}, \sqcup x_\alpha)$, where $x_\alpha$ is the basepoint. It seems that he is using that they form a good pair. However it does not seem to be necessarily true. Even if it is true, how does the claim follow from it?

koch
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    I think the result is not true if $(X_\alpha,x_\alpha)$ are not good pairs – Maxime Ramzi Feb 19 '19 at 23:32
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    If they are good pairs then $\bar H_n(\vee X_\alpha)= H_n(\sqcup X_\alpha, \sqcup x_\alpha)$. Then $H_n(\sqcup X_\alpha, \sqcup x_\alpha)$ splits into a direct sum $\oplus H_N(X_\alpha,x_\alpha)$ where each summand is a good pair, so you get $\oplus \bar H_n(X_\alpha)$. – Connor Malin Feb 19 '19 at 23:38

1 Answers1

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You seem to be referring to Hatcher's Corollary 2.25. In the book, he explicitly states that it is assumed that each $(X_\alpha, x_\alpha)$ is a good pair.

As for the claim itself, the steps are: \begin{align} \widetilde H_n (\vee_\alpha X_\alpha ) & \cong H_n (\vee_\alpha X_\alpha, x_0 ) \\ &\cong H_n (\sqcup_\alpha X_\alpha / \sqcup_\alpha x_\alpha, \ \sqcup_\alpha x_\alpha / \sqcup_\alpha x_\alpha ) \\ &\cong H_n (\sqcup_\alpha X_\alpha, \sqcup_\alpha x_\alpha) \\ &\cong \oplus_\alpha H_n (X_\alpha, x_\alpha) \\ &\cong \oplus_\alpha\widetilde H_n (X_\alpha)\end{align}

To spell it out:

  • First line: reduced homology is isomorphic to relative homology relative to a point.
  • Second line: the wedge sum is homeomorphic to the disjoint union quotiented by the disjoint union of the copies of the base points.
  • Third line: $H_n (X / A, A / A) \cong H_n (X, A)$ when $(X, A)$ is a good pair (Hatcher Proposition 2.22).
  • Fourth line: the homology of a disjoint union is the direct sum of the homologies of the components. Here, we're using a relative version of this fact.
  • Fifth line: reduced homology is isomorphic to relative homology relative to a point.
Kenny Wong
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