This question is about the following problem from Artin and a proof for the problem:
Prove that the nonempty fibres of a map form a partition of the domain.
Why is it not shown that the union of the fibres is the domain of the map? Or is this implicit from the definition of map?
Moreover (about the problem in general), don't we need to assume that the map is surjective? Otherwise, there would exist elements in the codomain such that its fibre is nonempty.
It's implicit in the definition of a map/function. If $f: X \to Y$ then every $x \in X$ is mapped to some element $f(x) \in Y$. So $x \in X$ but $x \not\in f^{-1}(y)$ for all $y \in Y$ would say that $f(x) \neq y$ for all $y$ and hence $f(x) \not\in Y$, which is impossible.
No, we don't need to assume surjectivity. I'm not sure I understand your point of confusion here. If we pick $y \in Y\setminus f(X)$, then $f^{-1}(y) = \{x \in X \mid f(x) = y\} = \emptyset$ is not a nonempty fibre. (So such fibres $f^{-1}(y)$ are not part of the partition under discussion.)
