Let $\mathbb F_q$ be a finite field of cardinality $q$ and let $A/\mathbb F_q$ be an abelian variety. Denote by $\pi_A:A\rightarrow A$ the Frobenius morphism which is defined by a $q$-th power map on the coordinates.
To compute the number of $\mathbb F_q$-points of $A$ we need to compute the kernel of $\pi_A-\text{id}$ on $A(\overline{\mathbb F}_q)$. For this, we look at $\ker(\pi_A-\text{id})$ as a scheme over $\mathbb F_q$. Next, we would argue that $\pi_A-\text{id}$ is separable and therefore $\#\ker(\pi_A-\text{id})=\deg(\pi_A-\text{id})$.
For the above to work we need in particular $\pi_A-\text{id}$ to be a finite morphism (an isogeny). However, since $\pi_A$ is the identity on the topological space $|A|$, it apears that the image of $\pi_A-\text{id}$ as a set should be a single point, meaning in particular that the morphism $\pi_A-\text{id}$ is not surjective and thus not finite.
Something is surely wrong in my argument but I cannot understand what am I missing.
The homomorphism $\pi_A$ acts identically on $\mathbb{F}_q$-points, but highly nontrivially on points which are defined over extensions of $\mathbb{F}_q$. As a homomorphism, $\pi_A$ and $\text{Id}$ are very different.
To see that $\pi_A - \text{Id}$ is finite, one can argue as follows. First, $\pi_A - \text{Id}$ is a homomorphism. Next, the induced map on tangent space is $-\text{Id}$. This is because $\pi_A$ acts trivially on tangent space.