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Question. Let $f : A \to B$ be a morphism between two abelians groups. Suppose that the induced morphism $\bar{f} : A \otimes_{\mathbb{Z}} \mathbb{R} \to B \otimes_{\mathbb{Z}} \mathbb{R}$ defined by $\bar{f}(a \otimes x) = f(a) \otimes x$ is an isomorphism. Is it necessarily the case that the induced morphism $A \otimes_{\mathbb{Z}} \mathbb{Q} \to B \otimes_{\mathbb{Z}} \mathbb{Q}$ is an isomorphism?

Motivation. In homotopy theory, a continuous map between based spaces $f : X \to Y$ is called a rational homotopy equivalence if $\pi_n(f) \otimes \mathbb{Q} : \pi_n(X) \otimes \mathbb{Q} \to \pi_n(Y) \otimes \mathbb{Q}$ is an isomorphism for each $n$. Similarly, it's called a real homotopy equivalence if $\pi_*(f) \otimes \mathbb{R}$ is an isomorphism. It's clear that a rational homotopy equivalence is a real homotopy equivalence: $(- \otimes_{\mathbb{Z}} \mathbb{Q}) \otimes_{\mathbb{Q}} \mathbb{R} \cong - \otimes_{\mathbb{Z}} \mathbb{R}$. Is the converse true?

Progress. When $A$ and $B$ are finitely generated abelian groups, the statement is true. Use the decomposition as a direct sum of a torsion group and a free abelian group, so that $A \otimes \mathbb{R} \cong \mathbb{R}^r$ and $B \otimes \mathbb{R} \cong \mathbb{R}^s$. Since there's an isomorphism, $r=s$, and the determinant $\det(f \otimes \mathbb{R})$ is nonzero; but this determinant is equal to $\det(f \otimes \mathbb{Q})$ (in fact it's an integer, equal to $\det(f)$), so $f \otimes \mathbb{Q}$ is also invertible.

I haven't been able to generalize that to abelian groups that are not finitely generated. I suspect the answer is that it's false, but I couldn't come up with a counterexample (surjectivity seems to be the main problem).

Najib Idrissi
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  • What you're asking is whether $\mathbb{R}$ is faithfully flat over $\mathbb{Q}$, don't you? Hint: $\mathbb{Q}\subset \mathbb{R} $ is a direct summand (as $\mathbb{Q}$-vector spa e). – Ben Oct 06 '15 at 07:50
  • Let me say that if $R\subset S$ is a ring extension and $R$ is a direct summand of $S$ (as an $R$-module), then $S$ is not necessarily $R$-flat. (Of course, this is not the case for field extensions.) – user26857 Oct 06 '15 at 09:26
  • Dear @user26857, you're right. My hint was meant to point towards the faithfully. Of course, arguing via freeness as you did is quicker... – Ben Oct 06 '15 at 15:04
  • By the way, if anyone is reading this: a real homotopy equivalence is not what I wrote in my question. In general it's simply a morphism between the real minimal models of the spaces; it doesn't have to induce a continuous map! – Najib Idrissi Apr 14 '17 at 12:16

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$\mathbb R$ is a faithfully flat $\mathbb Q$-module (since it is free). Then the sequence $0\to A\otimes_{\mathbb Z}\mathbb Q\to B\otimes_{\mathbb Z}\mathbb Q\to0$ tensorized by the $\mathbb Q$-module $\mathbb R$ is exact, so it is exact.

user26857
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