In Algebraic Varieties, Kempf proves a slight generalization of Chow's lemma (in the case of varieties): letting $Y$ be an irreducible separated variety, there exists an irreducible projective variety $X$ and a closed subset $Z$ of $X \times Y$ such that $\pi_X$ is an isomorphism onto an open subset and $\pi_Y$ is birational.
In his proof he covers $Y$ by open affines $Y_1, \ldots, Y_n$ and takes $O = \bigcap Y_i$. Then he embeds each $Y_i$ as an open subset of a projective variety $\overline{Y_i}$ and takes the map $O \to O \times \ldots \times O \to Y_1 \times \ldots \times Y_n \to \overline{Y_1} \times \ldots \times \overline{Y_n}$, where the first map is the diagonal and the second and third are inclusions in each coordinate. Then he lets $X$ be the closure of $O$ in $\prod \overline{Y_i}$, and $Z$ be the closure of the graph of the inclusion $O \to Y$ in $X \times Y$. It is pretty easy to see that $\pi_Y$ is an isomorphism on $O$ embedded in $X$, hence is birational.
To show that $\pi_X$ is an isomorphism (this is where the proof seems to diverge from the common one for Chow's lemma) he lets $Z_i = Z \cap (X \times Y_i)$ and tries to write each $Z_i$ as the graph of a morphism $U_i \to Y_i$ with $U_i$ open in $X$. He lets $U_i$ be $X \cap (\overline{Y_1} \times \ldots \times Y_i \times \ldots \times \overline{Y_n})$ and writes $f_i \colon U_i \to Y_i$ as the projection. It is easy to see that $Z_i$ contains the graph of $f_i$. His proof that the graph of $f_i$ contains $Z_i$ is where I am confused.
This is equivalent to $(y_1, \ldots, y_n, y) \in Z_i$ implying $y = y_i$, or $Z_i \subseteq V(y - y_i)$. What he writes is, verbatim, "This is trivial because the transposed graph $(y, y)$ of the inclusion of $Y_i$ in $\overline{Y_i}$ is closed in $\overline{Y_i} \times \overline{Y_i}$, and is the closure of $\{(u, u) \mid u \in O\}$."
I understand why this claim is true, but I don't see why it implies that $Z_i$ is contained in $V(y - y_i)$. The closure of $\{(u, u) \mid u \in O\}$ in $X \times Y$ will be $Z$, and maybe you can express $Z_i$ as the closure of $\{(u, u) \mid u \in O\}$ in $X \times Y_i$, so so that $\{(u, u) \mid u \in O\}$ implies $Z_i \subseteq V(y - y_i)$ in $X \times Y_i$, but I'm not sure that that is true. Any help is appreciated!
After thinking about it for a few days I think the idea is that the closure of $\Delta_O$ in $X \times Y_i$ is of course the intersection of the closure of $\Delta_O$ in $X \times Y$ with $X \times Y_i$, that is, $Z \cap X \times Y_i = Z_i$. Now since the inclusion map of both $\Delta_O$ and the transposed graph into $X \times Y_i$ is the diagonal and $Y$ is separated it should be closed, and hence it preserves closures, so that $Z_i$ is the image of the transposed graph. This image is clearly contained in $V(y - y_i)$ so we are done. I'm not positive this is correct but it seems right to me.