How to construct a homomorphism from an abelian group to $\mathbb{Q}/\mathbb{Z}$? Let $A$ be an abelian group. 
I don't know how to construct a nontrivial map from $A$ to the group $\mathbb{Q}/\mathbb{Z}$.
Can anyone help me, please?
 A: $\mathbb{Q}/\mathbb{Z}$ is a divisible module over the PID $\mathbb{Z}$, so it is injective.  (For definitions and proofs, see e.g. $\S$ 3.6 of these notes.)
Injectivity means: if $B$ is a submodule of $A$, then every homomorphism $f: B \rightarrow \mathbb{Q}/\mathbb{Z}$ extends to a homomorphism $f: A \rightarrow \mathbb{Q}/\mathbb{Z}$.  Since $A$ is nontrivial, it has a nontrivial cyclic subgroup $B$.  Defining a nontrivial homomorphism from a cyclic $\mathbb{Z}$-module to $\mathbb{Q}/\mathbb{Z}$ is easy: I leave it to you to find all such homomorphisms.
Added: I just checked Weibel's book.  The facts I mentioned in the first paragraph above are Corollary 2.3.2 on p. 39.  The OP's question is an exercise on p. 40.  So I have little doubt that this is the intended solution.
A: Let $G=\mathbb Q/\mathbb Z$, and suppose $A\neq0$.
Consider the set $S$ of pairs $(B,\phi)$ with $B$ a non-zero subgroup of $A$ and $\phi:B\to G$ a non-zero homomorphism. The set $S$ is not empty: if $a\in A$ is any non-zero element, then the cyclic subgroup $(a)$ is non-zero and there is a map $\phi:(a)\to G$, as one easily sees.
There is a partial order on $S$, as follows: $(B_1,\phi_1)\preceq(B_2,\phi_2)$ iff $B_1\subseteq B_2$ and $\phi_2|_{B_1}=\phi_1$. One can easily check that the partially ordered ser $(S,\preceq)$ satisies the hypotheses of Zorn's lemma. There exists then a maximal element $(A_0,\phi_0)$.
Suppose $A_0\subsetneq A$, so that there is an element $x\in A\setminus A_0$.
There exists $n\in\mathbb N_0$ such that the subgroup $A_0\cap (x)$ of $(x)$ is generated by $nx$.
Let us suppose first that $n\neq0$, and set $y=\tfrac1n\phi_0(nx)\in G$. I claim that there is an homomorphism $\phi:A_0+(x)\mapsto G$ such that whenever $a\in A_0$ and $t\in\mathbb Z$ we have $\phi(a+tx)=\phi_0(a)+ty$. To check this,  we first need to show that this is well defined; do that and then check that we have an homomorphism.
Now $(A_0+(x),\phi)$ is an element of $S$ strictly larger than $(A_0,\phi_0)$! This is absurd, and therefore we must have $A=A_0$, so that $\phi_0$ is a homomorphism from $A$ to $G$.
we are left with the case in which $n=0$, which is easier. Try it :-)
A: Any abelian group $A$ can be embedded into a divisible group $Q$ (see for example this answer). Moreover, an abelian divisible group can be written as a direct sum of $\mathbb{Q}$ and Prüfer groups; so if $A$ is torsion-free (resp. a torsion group), we can suppose that $Q$ is also torsion-free (resp. a torsion group).
Let $A$ be a nontrivial abelian group and $T$ its torsion subgroup.
Case 1: Suppose $T=A$. Then $A \hookrightarrow \bigoplus\limits_{p \in K} \mathbb{Z}[p^{\infty}]$, and because $\mathbb{Q}/ \mathbb{Z} \simeq \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]$, we deduce that $A \hookrightarrow \bigoplus\limits_{i \in I} \mathbb{Q}/ \mathbb{Z}$. Now we can find a projection such that $A \to \mathbb{Q} / \mathbb{Z}$ is nontrivial.
Case 2: Suppose $T \neq A$. Then $A/T$ is a nontrivial abelian torsion-free group, hence $A/T \hookrightarrow \bigoplus\limits_{i \in I} \mathbb{Q}$ and we can find a nontrivial projection $A/T \to \mathbb{Q}$. So $$A \twoheadrightarrow A/T \hookrightarrow \bigoplus\limits_{i \in I} \mathbb{Q} \twoheadrightarrow \mathbb{Q} \twoheadrightarrow \mathbb{Q}/ \mathbb{Z}$$
If the image of $A$ in $\mathbb{Q}/ \mathbb{Z}$ is trivial, its image in $\mathbb{Q}$ is included into $\mathbb{Z}$, so it is sufficient to compose by the automorphism $x \mapsto \frac{1}{2} x \in \mathrm{Aut}(\mathbb{Q})$ so that the given morphism can be supposed not trivial.
