# Dominating morphism $f:X \longrightarrow Y \Rightarrow$? any irreducible component of $X$ dominates $Y$ [closed]

Let $f:X \longrightarrow Y$ be a morphism such that any irreducible component of $X$ dominates $Y$. $(1)$

My question is:

a) what does it mean any irreducible component of $X$ dominates $Y$?

b) if $f:X \longrightarrow Y$ is a dominant morphism, it's true $(1)$?

## closed as off-topic by Saad, Ethan Bolker, Namaste, JonMark Perry, user99914 Jun 17 '18 at 16:25

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a) It just means that if you restrict the map $f$ to an irreducible component, the restricted map is dominant.
b) No. Consider, $\operatorname{Spec}(k[x,y]/(xy))\rightarrow \operatorname{Spec}(k[x])$. You can see that the restriction to the "$x$-axis" dominates, but the restriction to the "$y$-axis" does not.