Lemma 95.18.3. The $1$-morphism $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X})$ is faithful.
Proof. To check that $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X})$ is faithful it suffices to prove that it is faithful on fibre categories. Suppose that $\xi = (U, Z, y, x, \alpha )$ and $\xi ' = (U, Z', y', x', \alpha ')$ are two objects of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over the scheme $U$. Let $(g, b, a), (g', b', a') : \xi \to \xi '$ be two morphisms in the fibre category of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$. The image of these morphisms in $\mathcal{H}_ d(\mathcal{X})$ agree if and only if $g = g'$ and $a = a'$. Then the commutative diagram
\[ \xymatrix{ y|_ Z \ar[rr]_\alpha \ar[d]_{b|_ Z, \ b'|_ Z} & & F(x) \ar[d]^{F(a) = F(a')} \\ y'|_ Z \ar[rr]^-{\alpha '} & & F(g^*x') = F((g')^*x') \\ } \]
implies that $b|_ Z = b'|_ Z$. Since $Z \to U$ is finite locally free of degree $d$ we see $\{ Z \to U\} $ is an fppf covering, hence $b = b'$. $\square$
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