Denote by $F_q$ the finite field with $q$ elements, and denote by $\bar{F_q}$ its algebraic closure. Let $V$ be an affine $\bar{F_q}$-variety and $F$ be the Frobenius endomorphism corresponding to an $F_q$-structure $V(F_q)$ on $V$.
Let $R$ be the ring of polynomial functions of $V$, and $R=R(F_q)⊗\bar{F_q}$. The Frobenius endomorphism on $V$ induces a ring homorphism $F^*$ from $R$ to $R$ given by $f\otimes\lambda\mapsto f^q\otimes\lambda$.
Now turn to the question. It seems that $F^*$ is not surjective. Right? However, there is a statement as following.
Let $V$ be an affine $\bar{F_q}$-variety with ring of polynomial functions $R$. A surjective morphism $R$ to $R$ is the Frobenius endomorphism attached to an $F_q$-structure on $V$ if and only if for any $x\in R$ there exists n such that $F^{*n}(x)=x^{q^n}$.
Hence, I wonder that how $F^*$ can be surjective in the statement.
Is $F^*$ really surjective always?
$F^*$ is not surjective in general. A counter-example is as following.
Let $V=\mathbb{A}^1$, and then $R=\bar{F_q}[T]=F_q[T]\bigotimes\bar{F_q}$. For each element $f(T)\otimes\lambda\in R$, we have $F^*(f(T)\otimes\lambda)=f(T)^q\otimes\lambda=f(T^q)\otimes\lambda$. It is obvious that $T\otimes1$ has no preimage.