$\mathbb{A}_{\mathbb{F}_p}^1\rightarrow \mathbb{A}_{\mathbb{F}_p}^1$ corresponding to $x\mapsto x^p$ induces an isomorphism on residue fields

Question

I want to prove that the morphism of affine schemes $\mathbb{A}_{\mathbb{F}_p}^1\rightarrow \mathbb{A}_{\mathbb{F}_p}^1$ defined by the ring homomorphism $\varphi :\begin{matrix}\mathbb{F}_p[x]\rightarrow \mathbb{F}_p[x]\\ f(x)\mapsto f(x^p) \end{matrix}$ is bijective as a map of sets and it induces an isomorphism on residue fields at closed points.
The non zero prime ideals of $\mathbb{F}_p[x]$ are of the form $(f)$ where $f$ is an irreducible polynomial. If $\mathfrak{p}=(f)$ is such a prime ideal then $\varphi^{-1}(\mathfrak{p})=\{g\in\mathbb{F}_p[x],\ f|g^p\}=\{g\in\mathbb{F}_p[x],\ f|g\}=(f)=\mathfrak{p}$ so the underlying map on topological spaces is just the identity map.
Now let $\mathfrak{p}=(f)$ be a closed point of $\mathbb{A}_{\mathbb{F}_p}^1.$ The residue field $\kappa(\mathfrak{p})$ at $\mathfrak{p}$ is $\mathbb{F}_p[x]_{\mathfrak{p}}/(f\mathbb{F}_p[x]_{\mathfrak{p}})$ and the induced homomorphism $\kappa(\mathfrak{p})\rightarrow \kappa(\mathfrak{p})$ is given by: $\frac{g}{h}\mapsto \frac{g^p}{h^p}$. I can't see why this morphism is surjective, any help would be appreciated !

Answer 1

Combining comments into an answer:

If $f$ has degree $n$, then elements of $\kappa(\mathfrak{p}) = \mathbb{F}_p[]/\mathfrak{p}$ can be represented by polynomials of degree $< n$, and there are only finitely many of those. Now field morphisms are always injective, and an injective map between finite sets of the same cardinality must also be surjective. Thus $\varphi$ is surjective.