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I have a question concerning example 3.14 of Hatcher's Algebraic Topology. There are a few details that confuse me, so I suppose the goal of my questions is more or less clarification - I apologize if they seem trivial. The first (ring)-isomorphism written is $$H^*(\coprod_\alpha X_\alpha;R)\stackrel{\cong}{\longrightarrow}\prod_\alpha H^*(X_\alpha;R)\quad(*)$$ Have I understood it correctly when I say that the argument is based on the fact that the inclusions $i:X_{\alpha_0}\hookrightarrow\coprod_\alpha X_\alpha$ induce, on the level of groups, the isomorphism $(*)$ and since continuous maps induce morphisms that respect the $\cup$-product, then $(*)$ is in addition in fact a ring-isomorphism?

If so, then I can not quite understand why we need he additional assumption on the pair $(X_\alpha,x_\alpha)$ when arguing that we have an isomorphism similar to $(*)$ for the wedge sum. Consider the following reasoning: On the level of groups we have $$H^*(\bigvee_\alpha X_\alpha;R):= \bigoplus_{i=0}^\infty H^i(\bigvee_\alpha X_\alpha;R)\stackrel{\prod i^*}{\to}\prod_\alpha\bigoplus_{i=0}^\infty H^i(X_\alpha;R)=\prod_\alpha H^*(X_\alpha;R)$$ where $\prod i^*$ is induced by the inclusion forming an isomorphism by the Eilenberg-Steenrod axiom for the wedge sum. Again, since continuous maps induce ring-morphisms, the isomorphism is in fact a ring-isomorphism and the claim follows. Of course something is wrong because I do not apply the additional assumption Hatcher makes. If anyone would care to point out where my mistake is and explain how one implement the assumption correctly, I would appreciate that very much. Thank you in advance!

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    $\begingroup$ Here's an example of where things go wrong without the good pair assumption. Without the assumptions it is the isomorphism that fails to exist, not that it preserve ring structures. Note that the second isomorphism you write down only holds for reduced cohomology. $\endgroup$
    – Tyrone
    Mar 8 at 14:02
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    $\begingroup$ Here is an even better example. In this paper (Th.1) the authors compute the first singular homology group of the Griffith's twin cone (it is isomorphic to the corresponding group of the harmonic archipelago). You can read about this space on Jeremy Brazas's blog. It is a non-contractible wedge of two contractible spaces. Here you can see how badly the wedge axiom for (co)homology can fail when the spaces involved do not have good basepoints. $\endgroup$
    – Tyrone
    Mar 8 at 15:49
  • $\begingroup$ @Tyrone, thanks a lot for the references! $\endgroup$
    – Ellefee
    Mar 9 at 9:22
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"is induced by the inclusion forming an isomorphism by the Eilenberg-Steenrod axiom for the wedge sum"

No it is not. The Eilenberg-Steenrod axioms in Hatcher are stated for CW-pairs, for which the "well-pointed axiom", that there is a neighborhood of $x_\alpha$ deformation retracting to it, is true for an arbitrary point in the CW complex.

The statement you want to prove relies on an argument precisely dual to the one for homology. In homology Hatcher's book states this assumption: see Corollary 2.25. It is the assumption that $(X, x_\alpha)$ is a "good pair".

If you want to state Eilenberg-Steenrod axioms outside the category of CW-pairs you need "good-pair" type assumptions.

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  • $\begingroup$ Alright, this makes sense, thank you for the reference. So the statement "is induced by the inclusion forming an isomorphism by the Eilenberg-Steenrod axiom for the wedge sum" is true once I assume that the pair $(X_\alpha,x_\alpha)$ is a good pair for all $\alpha$? $\endgroup$
    – Ellefee
    Mar 7 at 15:57
  • $\begingroup$ Sorry, maybe that question did not quite make sense. What I meant was, the induced morphism is an isomorphism once I assume that the pair $(X_\alpha,x_\alpha)$ is good? $\endgroup$
    – Ellefee
    Mar 7 at 15:59

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