Rigidity for a proper geometrically integral factor

Let $E$ be the closed equalizer of $f$ and $f(x_0,-)\circ {\mathrm{pr}}_Y$, obtained by pulling back the diagonal of $Z$. Write $p:X{\times }_{k}Y\to Y$ and $C=Y\setminus p((X{\times }_{k}Y)\setminus E)$. By [F3], $p$ is open, so $C$ is closed. A point $y$ belongs to $C$ exactly when the whole topological fibre of $p$ over $y$ lies in $E$. That fibre is geometrically integral and thus reduced, so the two morphisms agree on it scheme theoretically: the ideal of the equalizer vanishes at every point and is zero on a reduced scheme. In particular $y_0\in C$.

Fix $y\in C$. Its fibre has image the single point $f(x_0,y)$. Choose an affine open $W\subset Z$ containing that point. The closed subset $f^{-1}(Z\setminus W)$ has closed image under the proper projection $p$. Its image omits $y$, so an affine open neighbourhood $U=\mathrm{Spec}R$ of $y$ avoids that image. Consequently $f(X{\times }_{k}U)\subset W$. By [F1] and [F2], $\mathrm{\Gamma }(X{\times }_{k}U,\mathcal{O})=k{\otimes }_{k}R=R$. The map to the affine $W$ therefore factors through $U$, and evaluating at $x_0$ identifies the factor as $f(x_0,-){\mathrm{\mid }}_U$. Thus $U\subset C$, proving that $C$ is open.

Since $Y$ is connected and $C$ is nonempty, open, and closed, $C=Y$. The factorization in step 2.1 holds on an open neighbourhood of every point, hence glues to the asserted equality of morphisms on all of $X{\times }_{k}Y$. AC is carried from the proper global-functions and morphism suppliers; no stronger field or projectivity assumption is used.