Algebraic Topology · Problem 3F.9

Cohomology — Allen Hatcher

Mathematics — Topology · ext-divisible-torsionfree-splicing

Problem

For an abelian group AA let p:A→Ap: A \to A be multiplication by pp, and let pA=ker⁡p{}_{p}A = \ker p, pA=Im ppA = \mathrm{Im}\, p, and Ap=Coker pA_p = \mathrm{Coker}\, p as in the proof of Proposition 3F.12. Show that the six-term exact sequences involving Hom(−,Z)\mathrm{Hom}(-, \mathbb{Z}) and Ext(−,Z)\mathrm{Ext}(-, \mathbb{Z}) associated to the short exact sequences 0→pA→A→pA→00 \to {}_pA \to A \to pA \to 0 and 0→pA→A→Ap→00 \to pA \to A \to A_p \to 0 can be spliced together to yield the exact sequence across the top of the following diagram Hom(pA,Z)→Ext(Ap,Z)→Ext(A,Z)→pExt(A,Z)→Ext(pA,Z)→0\mathrm{Hom}({}_pA, \mathbb{Z}) \to \mathrm{Ext}(A_p, \mathbb{Z}) \to \mathrm{Ext}(A, \mathbb{Z}) \xrightarrow{p} \mathrm{Ext}(A, \mathbb{Z}) \to \mathrm{Ext}({}_pA, \mathbb{Z}) \to 0 where the map labeled 'pp' is multiplication by pp. Use this to show: (a) Ext(A,Z)\mathrm{Ext}(A, \mathbb{Z}) is divisible iff AA is torsionfree. (b) Ext(A,Z)\mathrm{Ext}(A, \mathbb{Z}) is torsionfree if AA is divisible, and the converse holds if Hom(A,Z)=0\mathrm{Hom}(A, \mathbb{Z}) = 0.

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