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I am working on the exercise below from an old commutative algebra qualifying exam. I've been studying Atiyah & Macdonald's commutative algebra text.

Let $\pi:\mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ be the projection map. Suppose that $A$ is a divisible abelian group, and suppose that $f,g:A \rightarrow \mathbb{Q}$ are group homomorphisms. Prove that, if $\pi \circ f = \pi \circ g$, then $f = g$.

We recall that an abelian group $A$ is called divisible if, for each positive integer $n$ and each element $a \in A$, there exists some element $\overline{a} \in A$ such that $n\overline{a} = a$.

The assumption $\pi \circ f = \pi \circ g$ tells us that we have the following commutative diagram:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} A & \ra{f} & \mathbb{Q} \\ \da{g} & & \da{\pi} & \\ \mathbb{Q} & \ra{\pi} & \mathbb{Q}/\mathbb{Z} \\ \end{array} $$

My idea was to augment this diagram into a commutative diagram with two exact sequences as its rows (for example, extending the bottom of this diagram to the exact sequence $0 \rightarrow \mathbb{Z} \xrightarrow{\iota} \mathbb{Q} \xrightarrow{\pi} \mathbb{Q}/\mathbb{Z} \rightarrow 0$), and then use a result such as the Snake Lemma or Five Lemma. But, I'm not sure how to proceed by using the assumption that $A$ is a divisible abelian group.

Any help would be appreciated. Thanks!

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If $h = f - g$, then $\pi \circ h = 0$ whence $\operatorname{im}(h)\subset \operatorname{ker}(\pi) = \mathbb{Z}$. If $a\in A$ and $n\in \mathbb{Z}^+$, use divisibility of $A$ to write $h(a) = n\, h(a')$ for some $a'\in A$. It follows that $h(a)$ is divisible by $n$ for every positive integer $n$, so $h(a) = 0$.

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