I'm stuck in a simple question about weak one way functions.
Suppose $f(x)$ is strong one way, is $g(x)=f(x)_0$, i.e. taking the first bit of $f(x)$ a weak one way function? Intuitively, it is, because for any adversary $\mathcal{A}$ the probability of inverting $g$ is $\frac{1}{2} + \epsilon$. So it seems safe to say $g$ is $\frac{2}{3}$-one way.
But apparently one cannot reduce inverting $g$ to inverting $f$, which leaves me no way to prove that statement by reduction. Am I getting it wrong?