# Commutative diagram involving divisible group

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!

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 $$h(a) = 0$$.