Let $f:X→Y$ be a proper + dominant map with connected fibres (i.e. $X_y$ connected topological space for all $y \in Y$) between integral proper schemes over an algebraically closed field $K$ of arbitrary characteristic. suppose $f$ has property
\begin{equation} \mathcal{O}_Y= f_* \mathcal{O}_X \tag{P} \end{equation}
now let $Y' \subset Y$ any (in general neither open nor closed) subset of $Y$ that can be endowed with reduced scheme structure compatible with canonical inclusion $i: Y' \to Y$. this induces fibre product $X' := X \times_Y \to Y'$ endowed with canonical projection map $p: X \times_Y Y' \to Y'$.
Q: is the property (P) preserved by induced canonical projection $p: X \times_Y \to Y'$? i.e. does $\mathcal{O}_{Y'} = p_* \mathcal{O}_{X'}$ hold?
I don't know if following approach make sense: we going to analyse the structure sheaf $\mathcal{O}_{X'}=\mathcal{O}_{X \times Y'}$. let $V \subset Y$ arbitrary open + affine subscheme and let $f^{-1}(V) = \bigcup_{i \in I} U_i$ and $i^{-1}(V) = \bigcup_{j \in J} W_j$ with $U_i \subset X, W_j \subset Y'$ open & affine. a prototype open affine set of $X'$ is $U_i \times _V W_i$ and $\mathcal{O}_{X'}(U_i \times _V W_i)= \mathcal{O}_X(U_i) \otimes_{\mathcal{O}_Y(V)} \mathcal{O}_{Y'}(W_j)$ by construction of fibre product.
observe that $\mathcal{O}_Y(V) \to \mathcal{O}_X(U_i) $ comes from composition $\mathcal{O}_Y(V) \to f_*\mathcal{O}_X(V) =\mathcal{O}_X(\bigcup_{i \in I} U_i)$ induced by $\mathcal{O}_Y \to f_*\mathcal{O}_X$ composed with restriction $\mathcal{O}_X(\bigcup_{i \in I} U_i) \to \mathcal{O}_X(U_i) $. analogously for $\mathcal{O}_Y(V) \to \mathcal{O}_{Y'}(W_i) $.
since $V$ was chosen arbitrary $Y'$ is covered by such $W_i$ if we run over all $V \subset Y$ open + affine. therefore in order to prove (P) we want to show now that
\begin{equation} \mathcal{O}_{Y'}(W_i)= p_* \mathcal{O}_{X'}(W_i) \tag{1} \end{equation}
holds. we have $p_* \mathcal{O}_{X'}(W_i)= \mathcal{O}_{X'}(p^{-1} (W_i))$. it is in general not true that $p^{-1} (W_i)= f^{-1}(Z) \times_Z W_i$ for certain affine open $Z \subset Y$. from here I don't know how to finish the proof.