Induced map on the 2-part of an abelian group

Question

Let $A,B$ be abelian groups. Let $A[2]=\{a\in A\mid 2a=0\}$. Let $f:A \to B$ be a surjective homomorphism. Let $f':A[2]\to B[2]$ be the induced group homomorphism. When is $f'$ surjective ?

In the case $A=A[2]$ holds, it is clearly surjective. But if $A\neq A[2]$, what kind of condition ensures $f'$ is surjective ?

Background: Let $K$ be a quadratic number field. There is surjective map from $Cl(K) \to Cl(K,S)$ where $Cl(K,S)$ is $S$-ideal class group (Relation between $S$-ideal class group and usual ideal class group). I want to know when $Cl(K)[2]\to Cl(K,S)[2]$ is surjective. Especially, I want to prove $Cl(K)[2]=0 \implies Cl(K,S)[2]=0$.

Answer 1

$ Cl(K)[2]=1$ implies $Cl(K)$ is odd abelian group. There is surjection $Cl(K) \to Cl(K,S)$, then order of $Cl(K,S)$ is odd. Thus $Cl(K,S)[2]=1$.