Let M be abelian group and N be its subgroup . Now suppose We want to show any Hom(N,Q/Z) restriction of some Hom(M,Q/Z).
Attempt: Define f: M→ Q/Z as f(x) if x belongs to N and 0 if it does not. Clearly this is homomorphisam for M but if we restrict it then how to show that for this restriction of Hom(M,Q/Z) represent any Hom(N,Q/Z)?
On
Let $f \in \mathrm{Hom}(N,\Bbb Q/\Bbb Z)$
Consider the following collection $\mathcal X:=\{(U,g)\mid N \subset U \subset M, U \text{ is a subgroup of } M, g \in \mathrm{Hom}(M,\Bbb Q/\Bbb Z),g_{\mid N} = f\}$ with the partial order given by $(U,g) \leq (U',g') :\Leftrightarrow U \subset U', g'_{\mid U}=g$
By Zorn's lemma $\mathcal X$ has maximal elements, let $(U,g)$ be a maximal element.
Suppose $U \neq M$, then there is some $x \in M \setminus U$.
If $ \forall n \in \Bbb N : nx \notin U$, then $U \cap \langle x \rangle = \{0\}$, so $U + \langle x \rangle = U \oplus \langle x \rangle $ and one may extend $g$ by defining $g'(u \oplus nx)=g(u)$, where $u \oplus nx\in U \oplus \langle x \rangle$.
If $\{n \in \Bbb N \mid nx \in U\}$ is non-empty, let $n_0:=\min \{n \in \Bbb N \mid nx \in U\}$ and let $q \in \Bbb Q/\Bbb Z$ such that $n_0q=g(n_0x)$, now define $g'(u+nx):=g(u)+nq$ for $u+nx \in U + \langle x\rangle$
It is not clear that $g'$ is well-defined. Suppose $u+nx=u'+n'x$, then $u-u' =(n'-n)x \in \langle x \rangle \cap U=\langle n_0x \rangle $, so $n'-n=kn_0$ for some $k \in \Bbb Z$, so $$g(u)-g(u')=g(u-u')=g(kn_0x)=kg(n_0x)=kq=n'q-nq \Rightarrow g(u)+nq=g(u')+n'q$$ which shows that $g'$ is well-defined. Using the well-definedness, it's clear that $g'$ is a homomorphism.
So in either case, one can finds that $(U,g) \leq (U + \langle x \rangle, g')$ which contradicts the maximality of $U$, thus $U=M$ and $g$ is the desired extension of $f$.
Use the proof of Baer's criterion, i.e. apply Zorn's lemma to the poset $\{(N', \varphi) \mid N \subseteq N' \subseteq M, \varphi \in \operatorname{Hom}(N',\Bbb Q/\Bbb Z), \varphi|_M = g\}$ where $g$ is your homomorphism $N \to \Bbb Q/\Bbb Z$.