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Consider six countable Abelian groups and six group homomorphims as in the following diagram

G → H → I
↑       ↓
L ← K ← J

Assume that the resulting sequence is exact at all six entries.

Question: Is there a (second countable) locally compact Hausdorff space X with a closed subspace A, such that the resulting six-term sequence in K-theory

K0(X,A) → K0(X) →  K0(A)
  ↑                 ↓
 K1(A)  ← K1(X) ← K1(X,A)

is isomorphic to the above one?

An answer in the finitely generated case would also be interesting.

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First, it's equivalent to ask whether the sequence comes from a cofibration $X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}\Sigma X$ of finite spectra, which seems more natural. It is illuminating to consider the case where all groups are $\mathbb{Z}/4$ and all maps are multiplication by 2. I suspect that this cannot be done. Certainly, we cannot take $Y=X$ and $f=2.1_X$ because then $Z=S/2\wedge X$ and $K\wedge S/2$ has exponent $2$ (even though $S/2$ does not) so $K_0(Z)$ cannot be $\mathbb{Z}/4$. We could try taking $f$ to be twice an Adams self-map, but that does not work either. – Neil Strickland Feb 22 '11 at 11:30
@Neil Strickland: Yes, this is the simplest example in which problems arise. For, if the given algebraic sequence splits as a direct sum of sequences in which at least one map vanishes, I can accomplish the realisation. – Rasmus Bentmann Feb 22 '11 at 15:11
Can any 6-term sequence be realized as a 6-term sequence coming from an extension of $C^*$-algebras? – Andreas Thom Feb 22 '11 at 16:08
@Andreas: Yes (provided that it is exact, of course). This is in Alexander Bonkat's thesis. – Rasmus Bentmann Feb 22 '11 at 17:00
The $C^*$-algebra case should imply the topological case, at least for finitely generated groups. Using $C^*$-algebras we get a cofibration sequence of $K$-module spectra. Any $K$-module spectrum with finitely generated homotopy is equivalent to $K\wedge X$, where $X$ is a wedge of terms $S^0$, $S^0/n$, $S^1$ or $S^1/n$. In this context it is not hard to calculate $[X,Y]$ and $[K\wedge X,K\wedge Y]^K$ and I think that with a bit of bookkeeping we can make things work out. It would probably be easier to adapt Bonkat's proof, however. What is his approach? – Neil Strickland Feb 22 '11 at 17:38

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