preimage of a torsion-free subgroup 
Let $\phi: G \to H$ be a surjective group homomorphism such that $\ker(\phi)$ is torsion-free. Let $B$ be a torsion-free subgroup of $H$. Show that $A = \phi^{-1}(B)$ is torsion-free.

I'm confused why we need the condition that $\ker\phi$ is torsion-free.

We let $a\not=e$ is torsion, so $\varphi(a)^n=\varphi(a^n)=e$. 
Now since $B$ is torsion-free, $\varphi(a)=e$, hence $a \in \ker \varphi$. This is a contradiction. 
