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  "body": "\n## Problem\n\nLet $A$ be an [[abelian group|082GHNPJ53]].\nThen $A$ is a *Whitehead group* if the following holds for every abelian group $B$.\nFor every surjective [[homomorphism|0JYTG5K9Y4]] $f \\colon B \\to A$ with kernel isomorphic to $\\Z$, there is a homomorphism $g \\colon A \\to B$ with $f \\circ g = \\id_A$.\nEquivalently, $\\mathrm{Ext}^1(A, \\Z) = 0$, in terms of the [[Ext functor|4P7GBWJ4RR]].\nThe Whitehead problem asks: is every Whitehead group a [[free abelian group|0WJX8MGG37]]?\n\n## Remarks\n\nEvery free abelian group is a Whitehead group, because a surjection onto a free abelian group always splits: send each basis element to any preimage.\nEvery countable Whitehead group is free, so the question concerns uncountable groups.\nShelah showed in 1974 that the problem is independent of [[ZFC|0HE5WDQ9FJ]].\nIf the [[axiom of constructibility|4QKRPRCSXH]] $V = L$ holds, then every Whitehead group is free.\nIf [[Martin's axiom|V8KGT8MRFE]] holds together with the negation of the continuum hypothesis, then there is a Whitehead group that is not free.\n\n## Explanations\n\n### L1 human\n\nAn abelian group is free if it has a basis, like $\\Z^n$.\nFree groups have a lifting property: any map onto them from a bigger group, with kernel $\\Z$, can be reversed.\nJ. H. C. Whitehead asked whether this lifting property characterises the free groups.\nThe surprising answer is that the usual axioms of set theory cannot decide it: in some models of set theory the answer is yes, in others it is no.\n",
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