Abstract
We prove that any open subgroup of the free abelian topological group on a completely regular space is a free abelian topological group. Moreover, the free topological bases of both groups have the same covering dimension. The prehistory of this result is as follows. The celebrated Nielsen–Schreier theorem states that every subgroup of a free group is free, and it is equally well known that every subgroup of a free abelian group is free abelian. The analogous result is not true for free (abelian) topological groups [1,5]. However, there exist certain sufficient conditions for a subgroup of a free topological group to be topologically free [2]; in particular, an open subgroup of a free topological group on a kω-space is topologically free. The corresponding question for free abelian topological groups asked 8 years ago by Morris [11] proved to be more difficult and remained open even within the realm of kω-spaces. In the present paper a comprehensive answer to this question is obtained.
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