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 $\sigma\in \mathrm{Gal}(X/Y)$.
I can´t follow the second part of the proof. Why is this true?
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 $X \times_Y X$.
