How to prove that existence of one-way functions implies P≠NP?

Wikipedia:

The existence of such one-way functions... would prove that the complexity classes P and NP are not equal.

How is this proved?

Suppose that P=NP, and that $f\colon \{0,1\}^ \to \{0,1\}^$ is an arbitrary function computable in polynomial time. Suppose that $|x| = n$, and we are given $y = f(x)$. We will show how to find $z \in \{0,1\}^n$ such that $y = f(z)$ in polynomial time (in $n$).

Using an NP oracle, we determine whether there exists $z$ such that $z_1 = 1$ and $y = f(z)$. If so, we set $w_1 = 1$, and otherwise, we set $w_1 = 0$. Using an NP oracle, we determine whether there exists $z$ such that $z_1 = w_1$, $z_2 = 1$, and $y = f(z)$. We set $w_2$ accordingly. Continuing in this way, we eventually found $z$ such that $y = f(z)$. Since we assumed that P=NP, this algorithm runs in polynomial time.

StackExchange Computer Science Q#139291, answer score: 5