If we take E to be an elliptic curve over $\mathbb{F}_q$ given by $$y^2=x^3+Ax+B$$ and $F_q$ to be the Frobenius endomorphism given by: $$F_q (x,y) = (x^q,y^q)$$ I am told that: $$\#E(\mathbb{F}_q) = deg(F_q-1)$$
In the very first line of the proof of why $\#(E(\mathbb{F}_q)) = \deg(F_q -1)$, it states that:
$$a \in E(\mathbb{F}_q) \Leftrightarrow F_q(a) = a$$
but I can't understand the backwards implication. Surely for any element $x \in \mathbb{F}_q$ we have that $x^q = x$, so it seems like $\forall a \in \mathbb{F}_q$, we should have that $F_q(a) = a$, not just for elements of $E(\mathbb{F}_q)$. What am I misunderstanding?
Isn't it obvious? For an element $x\in \bar{F_p}$, then we have $x^q=a$ if and only if $x\in F_q$. So $a\in E(\mathbb{F}_q)$ if and only if it is fixed by the action of Frobenius.