Let $X, Y$ be smooth projective connected complex varieties of the same pure dimension $d$ and $f : X\to Y$ a finite flat surjective morphism.


Let $\Delta\_X$ be the closed subscheme of $X\times X$ that is the scheme-theoretic image of the diagonal morphism $X\to X\times X$, and $\Delta\_Y$ the scheme-theoretic image of the diagonal morphism $Y\to Y\times Y$.



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> Do we have $f^{-1}\Delta\_Y=\Delta\_X$ scheme-theoretically? (i.e. $\Delta\_Y\times\_{Y\times Y}(X\times X)=\Delta\_X$)
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I'd expect $\Delta\_X$ to be one of the irreducible components of $f^{-1}\Delta\_Y$, but was wondering about an explicit example and whether one could say something about the discrepancy between $f^{-1}\Delta\_Y$ and $\Delta\_X$.



Finite flat pullback of the diagonal
If $f$ is finite flat of degree $d$, then $f \times f \colon X \times X \to Y \times Y$ has degree $d^2$, but $\Delta\_f \colon \Delta\_X \to \Delta\_Y$ has degree $d$. So equality cannot hold scheme-theoretically unless $f$ is an isomorphism.


A fairly explicit case is the finite étale Galois case, where $(f \times f)^{-1}(\Delta\_Y)$ is the disjoint union of the graphs $\Gamma\_\sigma$ of deck transformations $\sigma \colon X \to X$.


If $f$ is only étale but not Galois, then $(f\times f)^{-1}(\Delta\_Y)$ will be smooth and $\Delta\_X$ is still a connected component, but the other components will not map isomorphically onto $X$ under their projections.


If $f$ is generically étale (i.e. separable), then there is a dense open $U \subseteq Y$ above which the above holds, so $(f \times f)^{-1}(\Delta\_Y)$ is still generically smooth (i.e. geometrically reduced). But for each component of the branch divisor $D \subseteq Y$, there will be another irreducible component of $(f \times f)^{-1}(\Delta\_Y)$ intersecting $\Delta\_X$ above $D$.


In the inseparable case, you get situations where $\Delta\_X$ occurs with multiplicity $>1$, but that only happens in positive characteristic.


