Proof.
Let $y : \mathop{\mathrm{Spec}}(k) \to Y$ be a field valued point. Then the fibre $X_ y$ of $f$ at $y$ is connected by our assumption that $H^0(X_ y, \mathcal{O}_{X_ y}) = k$. Thus the rank of $\mathcal{E}$ is constant on the fibres. Since $f$ is open (Morphisms of Spaces, Lemma 67.30.6) and closed we conclude that there is a decomposition $Y = \coprod Y_ r$ of $Y$ into open and closed subspaces such that $\mathcal{E}$ has constant rank $r$ on the inverse image of $Y_ r$. Thus we may assume $\mathcal{E}$ has constant rank $r$. We will denote $\mathcal{E}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{E}, \mathcal{O}_ X)$ the dual rank $r$ module.
By cohomology and base change (more precisely by Derived Categories of Spaces, Lemma 75.25.4) we see that $E = Rf_*\mathcal{E}$ is a perfect object of the derived category of $Y$ and that its formation commutes with arbitrary change of base. Similarly for $E' = Rf_*\mathcal{E}^\vee $. Since there is never any cohomology in degrees $< 0$, we see that $E$ and $E'$ have (locally) tor-amplitude in $[0, b]$ for some $b$. Observe that for any $g : Y' \to Y$ we have $f'_*((g')^*\mathcal{E}) = H^0(Lg^*E)$ and $f'_*((g')^*\mathcal{E}^\vee ) = H^0(Lg^*E')$. Let $j : Z \to Y$ and $j' : Z' \to Y$ be the locally closed immersions constructed in Derived Categories of Spaces, Lemma 75.26.6 for $E$ and $E'$ with $a = 0$ and $r = r$; these are characterized by the property that $H^0(Lj^*E)$ and $H^0((j')^*E')$ are locally free modules of rank $r$ compatible with pullback.
Let $g : Y' \to Y$ be a morphism. If there exists an $\mathcal{N}$ as in the lemma, then, using the projection formula Cohomology on Sites, Lemma 21.50.1, we see that the modules
\[ f'_*((g')^*\mathcal{E}) \cong f'_*((f')^*\mathcal{N}) \cong \mathcal{N} \otimes _{\mathcal{O}_{Y'}} f'_*\mathcal{O}_{X'} \cong \mathcal{N}\quad \text{and similarly }\quad f'_*((g')^*\mathcal{E}^\vee ) \cong \mathcal{N}^\vee \]
are locally free of rank $r$ and remain locally free of rank $r$ after any further base change $Y'' \to Y'$. Hence in this case $g : Y' \to Y$ factors through $j$ and through $j'$. Thus we may replace $Y$ by $Z \times _ Y Z'$ and assume that $f_*\mathcal{E}$ and $f_*\mathcal{E}^\vee $ are locally free $\mathcal{O}_ Y$-modules of rank $r$ whose formation commutes with arbitrary change of base.
In this sitation if $g : Y' \to Y$ is a morphism and there exists an $\mathcal{N}$ as in the lemma, then the map (cup product in degree $0$)
\[ f'_*((g')^*\mathcal{E}) \otimes _{\mathcal{O}_{Y'}} f'_*((g')^*\mathcal{E}^\vee ) \longrightarrow \mathcal{O}_{Y'} \]
is a perfect pairing. Conversely, if this cup product map is a perfect pairing, then we see that locally on $Y'$ we have a basis of sections $\sigma _1, \ldots , \sigma _ r$ in $f'_*((g')^*\mathcal{L})$ and $\tau _1, \ldots , \tau _ r$ in $f'_*((g')^*\mathcal{E}^\vee )$ whose products satisfy $\sigma _ i \tau _ j = \delta _{ij}$. Thinking of $\sigma _ i$ as a section of $(g')^*\mathcal{L}$ on $X'$ and $\tau _ j$ as a section of $(g')^*\mathcal{L}^\vee $ on $X'$, we conclude that
\[ \sigma _1, \ldots , \sigma _ r : \mathcal{O}_{X'}^{\oplus r} \longrightarrow (g')^*\mathcal{E} \]
is an isomorphism with inverse given by
\[ \tau _1, \ldots , \tau _ r : (g')^*\mathcal{E} \longrightarrow \mathcal{O}_{X'}^{\oplus r} \]
In other words, we see that $(f')^*f'_*(g')^*\mathcal{E} \cong (g')^*\mathcal{E}$. But the condition that the cupproduct is nondegenerate picks out a retrocompact open subscheme (namely, the locus where a suitable determinant is nonzero) and the proof is complete.
$\square$
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