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The really beautiful way to prove the Universal Coefficients theorem, to my taste, is to use the fibration sequence $K(\mathbb{Z}, n) \to K(\mathbb{Z}, n) \to K(\mathbb{Z}/k, n)$ (I'm using $\mathbb{Z}/k$ coefficients for simplicity) to give a long exact sequence of cohomology, and split it into short exact sequences. (I'm pretty sure I learned this from Adams' Generalized Homology...).

Anyway, this gives the desired exact sequence, but I don't know how to get the (nonnatural) splitting that the homological algebra derivation provides.

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your proof yields the theorem only for CW-complexes, right? –  Martin Brandenburg Jun 13 '10 at 6:55
    
Well, it gives it for all spaces for which the cohomology theory you want is represented by maps into an Eilenberg-Mac Lane space. –  Jeff Strom Jun 13 '10 at 16:58

2 Answers 2

I would claim that the splitting (and indeed the whole universal coefficient theorem) is not really a topological theorem. If we take the homological version one really works with the chain complex $C_\ast(X)$ in the derived category of $\mathbb Z$-complexes. We then have $C_\ast(X,M)=C_\ast(X)\bigotimes M$ but as $C_\ast(X)$ is free this equals the derived tensor product $C_*(X)\bigotimes^{\mathbb L} M$ and hence is a formula in the derived category. One can then use the fact that in the derived category of $\mathbb Z$-modules every complex is isomorphic to the sum of its (shifted) homology: $C\cong \bigoplus_nH_n(C)[n]$ so that $$ C_*(X)\bigotimes^{\mathbb L} M \cong \bigoplus_n(H_n(X)\bigotimes^{\mathbb L} M)[n] $$ and as $A\bigotimes^{\mathbb L} M\cong A\bigotimes M\bigoplus \mathrm{Tor}^1(A,M)[1]$ we get the universal coefficient formula including the splitting.

This idea also demonstrates why the splitting is not canonical. We may for instance consider a group $G$ acting on $X$. We then get at complex $C_*(X)$ in the derived category of $G$-modules and a complex in that category is in general not isomorphic to the sum of its homology.

On the other hand, this technique can be used (essentially) each time some invariant of a topological space $X$ only depends on its chain complex in a way that takes quasi-isomorphisms to isomorphisms. The conclusion is that it only depends on the homology of $X$. A nice example is the homology of the $n$'th symmetric product of $X$. It turns out to be the homology of a complex constructed functorially from $C_*(X)$ and exactly in a way that preserves quasi-isomorphisms. Hence it only depends on the homology of $X$ (and one can also give explicit formulas).

However, the method that you declare a fondness for is also useful if one goes beyond homology. It can be used to give a universal coefficient spectral sequence (due to Adams I think) for the (co)homology with coefficients in module spectrum over a ring spectrum. In general this spectral sequence does not degenerate to short exact sequences so the problem of splitting is not even (that) relevant. However, for for instance $K$-theory it does but I imagine (though I don't know but others certainly do) that even there one can find examples of non-splitting.

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You seem to be suggesting that we you don't use chain complexes, we can't prove the splitting. –  Jeff Strom Jun 13 '10 at 17:04
    
I at least don't know of any way to do it without it. –  Torsten Ekedahl Jun 13 '10 at 17:26
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Since the splitting is not natural, you won't be able to see it by looking at the Eilenberg-Maclane spaces alone. If you could find the splitting as a map between them it would induce a natural transformation. I agree with Ekedahl, that the only way I see to get the spliting is to realize that a map from X into K(A,n)s factors through $X \wedge H \mathbb{Z} $, and that this is quasi-isomorphic to a sum of K(A,n)s in the category of HZ-module spectra. But this is just a fancy way of saying you need to use the chain complex to see the splitting. –  Chris Schommer-Pries Jun 14 '10 at 14:12
    
I think Chris' view point is sufficiently different from the one I presented to merit mention. I however also agree that it is conceptually very close: In my case a map (of simplicial sets says) $X \to K(A,n)$ which by adjunction is the same as $\mathbb Z[X] \to K(A,n)$ and we have a map of abelian simplicial groups, i.e., of chain complexes. In Chris case we have a (stable) map $X \to K(A,n)$ which by adjunction is the same as $X\wedge H\mathbb Z \to K(A,n)$, a map of $H\mathbb Z$-module spectra. (cont'd) –  Torsten Ekedahl Jun 14 '10 at 17:29
    
(cont'd) The end game is then the same; the derived categories of $\mathbb Z$-modules and of $\mathbb Z$-modules are (I hope) equivalent. –  Torsten Ekedahl Jun 14 '10 at 17:30

(This is a corrected version of my original, off-the--cuff answer).

If R has proj dim 1 and C is a flat chain complex, then you get the UCT sequence. To get the splitting, you need to also assume that C is projective, so that the map C -->> B from the complex to the subcomplex of boundaries has a splitting.

A spectrum level construction of the splitting seems unlikely since you don't have kernels and cokernels, but only fibers and cofibers there.

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Nope: just need R to be hereditary (hom dim le 1). Then, assuming C is a projective complex, the boundaries B are also projective and the map C -->> B splits. That's all that really matters. What I inaccurately remembered was that the UCT ses holds for flat complexes C, but that the proof that it splits that I know, only works if C is projective. So, if you can construct a flat complex C with C -->> B non-split you will probably have a nonsplit ses. This is all homological algebra -- nothing to do with homology of spaces or spectra. But this suggests why ... ack! 27 chars left. Bye. –  Robert Bruner Jun 17 '10 at 20:55
    
You should be able to edit your answer, just click on "edit" at the lower left (next to "link", "flag", and "delete"). There is no character limit for answers. –  Victor Protsak Jun 17 '10 at 23:49
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The abstract reason why this works is that you have a $t$-structure on a triangulated category. You then try to build up a general object by starting with shifts of objects in the heart and use iterated mapping cones. This gives all objects that have finite amplitude. At each stage the result mapping cone is classified by a suitable morphism. If $\mathrm{Hom}^i(M,N)=0$ for $i\geq2$ and $M$ and $N$ in the heart the result is that all maps are zero and any object of finite amplitude is the sum of shifts of its homology. –  Torsten Ekedahl Jun 19 '10 at 13:42
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(cont'd) For $R$-complexes this condition is global dim $\leq1$ but it is also fulfilled for $H(Z)$-module spectra. –  Torsten Ekedahl Jun 19 '10 at 13:42

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