 Solution

Assume $L^p\subseteq K$. For $b\in B$, its power $b^p$ belongs to $K$ and is integral over $A$, because $B$ is finite over $A$. Because $A$ is an integrally closed domain, $b^p\in A$. Thusis a ring homomorphism, defining a morphism $g_V:V\to f^{-1}V$. Its composition with $f:f^{-1}V\to V$ is the Absolute Frobenius morphism of $f^{-1}V$, because the composite ring map sends $b$ to $b^p$ in $B$.

These local morphisms glue. Explicitly, over a distinguished open $D(a)\subseteq V$, the ring rule isso it is exactly the localization of ${\gamma }_V$. For the underlying topology, each fibre of $f$ consists of a single point: if primes ${\mathfrak{q}}_1,{\mathfrak{q}}_2\subset B$ contract to the same $\mathfrak{p}$, thenso ${\mathfrak{q}}_1={\mathfrak{q}}_2$. The map $g_V$ sends that point of $V$ back to its unique point above it. Thus localization compatibility on a basis of a topology also gives compatibility on overlaps of arbitrary affine charts. We obtain

For uniqueness, $f$ is surjective on points, so the equation determines the point map of $g$. On each affine chart it then requires $f^{\ast }(g^{\ast }(b))=b^p$. The inclusion $A\hookrightarrow B$ is injective, so this determines $g^{\ast }(b)$ uniquely as $b^p$. Hence the sheaf map and the morphism are unique.