Let $f \colon X\to Y$ be a finite map of normal varieties. Then every irreducible component of $X\times\_{Y}X$ dominates $Y$.


The proof goes as follows. We can suppose that $f \colon X\to Y$ is a Galois covering. Then every irreducible component of $X\times\_{Y}X$ is the image of $f\_{\sigma}:X\to X\times\_{Y}X$, given by $f\_{\sigma}(x)=(x,x^{\sigma})$ for $\sigma\in \mathrm{Gal}(X/Y)$.



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> I can´t follow the second part of the proof. Why is this true?
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Finite map of normal varieties
By definition we have $$X \times\_Y X =\{(x\_1, \,x\_2) \in X \times X \; | \; f(x\_1)=f(x\_2)\}$$ and, since we are assuming that $f \colon X \to Y$ is a Galois morphism (this is not restrictive, because otherwise we can pass to the Galois closure), we can rewrite this as $$X \times\_Y X= \{(x\_1, \, x\_2) \; | \; x\_1 = x\_2^{\sigma} \; \; \textrm{for some} \; \, \sigma \in \mathrm{Gal}(X/Y) \}.$$ Then
$$X \times \_Y X = \bigcup\_{\sigma \in \mathrm{Gal}(X/Y)} X\_{\sigma},\; \; \textrm{where} \; \; X\_{\sigma}=\{(x, \, x^{\sigma}) \; | \; x \in X\}.$$


Each $X\_{\sigma}$ is irreducible, because it is the image of the irreducible variety $X$ via the morphism $f\_{\sigma}:X\to X\times\_{Y}X$ given by $f\_{\sigma}(x)=(x, \,x^{\sigma})$. Since the decomposition of a variety into irreducible components is unique, we conclude that the $X\_{\sigma}$ are precisely the irreducible components of $X \times\_Y X$. 


