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arXiv · 2102.04045

On the cokernel of the Baumslag rationalization

Sergei O. Ivanov
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Abstract

We prove that for the free group of rank two $F$ the cokernel of the homomorphism to its Baumslag rationalization $F\to {\sf Bau}(F)$ is not abelian. Moreover, this cokernel contains a free subgroup of countable rank. This answers a question of Emmanuel Farjoun.

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Sergei O. Ivanov. 2021-02-08. On the cokernel of the Baumslag rationalization. https://arxiv.org/abs/2102.04045

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