TITLE: Splitting of the Universal Coefficients sequence
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: (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.