A finite morphism $f : X \rightarrow Y$ between varieties (variety can be covered by a finite number of affine varieties) by definition is one where we can choose an open cover $U_i$ of $Y$ consisting of affines so that their inverse images are also affine and $k[f^{-1}U_i]$ is a finite $k[U_i]$ module through the induced map.
Now, in order to prove that the image is a closed set, author says it is a local property on $Y$, but I don't get how $im(f) \cap U_i$ being closed in $U_i$ subspace topology proves that $im(f)$ is closed in $Y$.
I am using Kempf's algebraic varieties.
Adding the answer here. $Y-f(X) = Y \cap f(X)^c = \cup_i (U_i \cap f(X)^c)$
is open since if each of $U_i \cap f(X)^c$ is open.