= 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>, $b^p\in A$. Thus
$$
\gamma_V:B\longrightarrow A,\qquad b\longmapsto b^p
$$
is 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 is
$$
\gamma_{D(a)}\left(\frac{b}{a^n}\right)=\frac{b^p}{a^{np}},
$$
so 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 $\mathfrak p$, then
$$
b\in\mathfrak q_j\iff b^p\in\mathfrak p,
$$
so $\mathfrak q_1=\mathfrak q_2$. 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
$$
\boxed{g:Y\longrightarrow X,\qquad g\circ f=F_X.}
$$
For uniqueness, $f$ is surjective on points, so the equation determines the point map of $g$. On each <affine chart> it then requires $f^*(g^*(b))=b^p$. The inclusion $A\hookrightarrow B$ is injective, so this determines $g^*(b)$ uniquely as $b^p$. Hence the <sheaf> map and the morphism are unique.
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