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Which of the statements is wrong:

  1. a generalized cohomology theory (on well behaved topological spaces) is determined by its values on a point
  2. reduced complex $K$-theory $\tilde K$ and reduced real $K$-theory $\widetilde{KO}$ are generalized cohomology theories (on well behaved topological spaces)
  3. $\tilde K(*)= \widetilde{KO} (*)=0$

But certainly $\tilde K\neq \widetilde{KO}$.

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1 is doubly wrong. First, you need to distinguished generalized cohomology theories and reduced generalized cohomology theories. If you want to work with the latter, you should replace "a point" in 1 by "$S^0$", and then the corrected version of 3 no longer holds. But even this new version 1' is false; a generalized cohomology theory is not determined by its coefficients, unless they are concentrated in a single degree (example: complex K-theory vs. integer cohomology made even periodic).

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  • $\begingroup$ Thanks, Reid. Someone should change the "Generalized cohomology theories" section in Wikipedia's article on "Cohomology". $\endgroup$
    – roger123
    Commented Mar 17, 2010 at 19:14
  • $\begingroup$ Hmm, I see. Well, Wikipedia does use quotes around "determined by its values on a point" :) but I agree that this statement is misleading at best. $\endgroup$
    – Reid Barton
    Commented Mar 17, 2010 at 19:24
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    $\begingroup$ It could easily be fixed by putting a link to the entry for the Atiyah-Hirzebruch spectral sequence as a way of making this precise. $\endgroup$
    – Andy Putman
    Commented Mar 17, 2010 at 19:27
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    $\begingroup$ We're thinking of generalized cohomology theories as taking values in graded abelian groups. Thus $\tilde{K}^q(S^0)\not\approx \tilde{KO}^*(S^0)$, for instance if $q\equiv -1,-2,-6\mod 8$. $\endgroup$
    – Charles Rezk
    Commented Mar 17, 2010 at 19:51
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    $\begingroup$ But the statement about cohomology theories being determined by their coefficients is not totally wrong -- if you have a natural transformation $K \to L$ of homology theories which induce an isomorphism on the point (or $\mathbf{S}^0$ in the case of reduced theories), then it's an isomorphism. But you do need to have a natural transformation in the first place. $\endgroup$
    – Tilman
    Commented Mar 17, 2010 at 22:08

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