Let $M$ be an abelian group. Every integer $n$ defines an endomorphism $f_n\colon M \rightarrow M$ such that $f_n(x) = nx$. $M$ is called divisible if $f_n$ is surjective for every nonzero integer $n$. Clearly $\mathbb{Q}/\mathbb{Z}$ is divisible.
Lemma 1 Let $M$ be a divisible abelian group. Let $I$ be an ideal of $\mathbb{Z}$. Then the canonical homomorphism $Hom(\mathbb{Z}, M) \rightarrow Hom(I, M)$ induced by the canonical injection $I \rightarrow \mathbb{Z}$ is surjective.
Proof: If $I = 0$, the assertion is clear. Hence we assume $I \neq 0$. There exists a nonzero integer $n$ such that $I = \mathbb{Z}n$. Let $f \in Hom(I, M)$. Since $M$ is divisible, there exists $a \in M$ such that $f(n) = na$. Let $x \in I$. There exists an integer $m$ such that $x = mn$. $f(x) = f(mn) = mf(n) = mna = xa$. Hence $f$ is in the image of the map$\colon Hom(\mathbb{Z}, M) \rightarrow Hom(I, M)$. QED
Lemma 2 Let $T$ be a divisible abelian group. Let $M$ be an abelian group. Let $N$ be a subgroup of $M$. Let $f\colon N \rightarrow T$ be a homomorphism. Let $x \in M - N$. Then there exists a homomorphim $g\colon N + \mathbb{Z}x \rightarrow T$ extending $f$.
Proof: Let $I = \{a \in \mathbb{Z}\colon ax \in N\}$. Let $h\colon I \rightarrow T$ be the map defined by $h(a) = f(ax)$. Since $h$ is a homomorphism, by Lemma 1, there exists $z \in T$ such that $h(a) = az$ for all $a \in I$.
Suppose $y + ax = y' + bx$, where $y, y' \in N, a, b \in \mathbb{Z}$. $y - y' = (b - a)x$ Hence $b - a \in I$. Hence $h(b - a) = f((b - a)x) = (b - a)z$. Hence $f(y - y') = (b - a)z$. Hence $f(y) + az = f(y') + bz$. Therefore we can define a map $g\colon N + \mathbb{Z}x \rightarrow T$ by $g(y + ax) = f(y) + az$. Clearly $g$ is a homomorophism extending $f$. QED
Theorem Let $T$ be a divisible abelian group. Then $T$ is injective.
Proof: This follows immediately from Lemma 2 and Zorn's lemma.
Corollary $\mathbb{Q}/\mathbb{Z}$ is injective.