Denote $R$ as the real number and $Z\subset R$ as integers. The purpose is to see $R/Z$ is injective.
This is from an exercise from Rotman 3.19 Homological Algebra.
Let $G$ be an abelian group and $a\in G$ be non-zero element. Prove there is a $f:G\to R/Z\cong S^1$ with $f(a)\neq 0$(in $S^1$, $f(a)=1$.
Replace $R/Z$ by $S^1$. If $a$ is non-finite order, I simply send $a$ to any non finite order element of $S^1$. If $a$ is finite order, then send $a$ to a element of $S^1$ with same order. So I obtain a map $f:<a>\to S^1$.
It suffices to extend $f:<a>\to S^1$ to $f:<a,b>\to S^1$ for any $b\not\in<a>$. So consider $\bar{b}\in\frac{<b,a>}{<a>}$. If $b$ is of finite order, there will be a $k\in Z$ such that $kb\in <a>$, take any $k-th$ root of $f(kb)$ to define $f'(b)$(i.e. $f'(kb)=f(kb)$). Then one obtain a map $f':<a,b>\to S^1$ by $f'(ka+lb)=f(a)^kf'(b)^l$. Well definedness can be checked. If $b$ is of infinte order, simply choose any other infinite order element as $f'(b)$ of $S^1$ distinct from $f(a)$. Then we define the map $f'(ka+lb)=f(a)^kf'(b)^l$. Since $kb$ never lies in $<a>$ for any $k\in Z$, the map is well define.
To see injectivity R/Z, simply extend the map and use zorn lemma(in analogy to the proof of Bayer criterion). Or the other way to see $R/Z$ is the injective envelope of $Q/Z$ which I am not sure as it need completion somewhere.
Am I on the right track for this problem? And I feel my extension of map part fishy.
Any divisible abelian group is injective (this holds more generally for modules over a principal ideal domain). Any quotient of a divisible group is divisible.
Since you mention Baer's criterion, let's use it.
Suppose $G$ is a divisible group and let $f\colon I\to G$ be a homomorphism, where $I$ is an ideal of $\mathbb{Z}$. If $I=\{0\}$, there's nothing to prove. Otherwise $I=n\mathbb{Z}$, with $n>0$. Let $x=f(n)$ and take $y\in G$ with $ny=x$ (it exists because $G$ is divisible). Defining $g\colon\mathbb{Z}\to G$ by $g(m)=my$ is the required extension of $f$.