Does "local on the target" mean the same thing as "local on the base"? In Vakil's book (Foundation of Algebraic Geometry), I learned the definition of a type of morphism of schemes being local on the target: 

A class of morphisms of schemes is called local on the target if 
(a) if $\pi:X\rightarrow Y$ is in the class then for any open set $V$
  of $Y$, the restricted morphism $\pi ^{-1} (V)\rightarrow V$ is in the
  class. 
(b) for a morphism $\pi:X\rightarrow Y$, if there is an open cover
  {$V_i$} of $Y$ for which each restricted morphism $\pi ^{-1}(V)\rightarrow V$ is in the class, then $\pi$ is in the class.

In the other book (Algebraic Geometry and Arithmetic curve Qing Liu), I learned a similar definition.

A property P of morphisms of schemes $\pi:X\rightarrow Y$ is said to
  be local on the base if the following assertions are equivalent:
(i) $\pi$ verifies P.
(ii) for any $y\in Y$, there exists an affine neighborhood $V$ of $y$
  such that the restricted morphism $\pi ^{-1} (V)\rightarrow V$ verfies
  P.

It's obvious that "local on the target" implies "local on the base", but do we have another direction?
 A: Local on the base implies condition (b), but not condition (a), of local on the target. 
For (b): Suppose $(V_i)_{i\in I}$ is an open cover of $Y$ such that $\pi_i\colon \pi^{-1}(V_i)\to V_i$ satisfies property $P$. By (i) implies (ii), for all $i\in I$ and all $y\in V_i$, there is an affine neighborhood $U_{i,y}$ with $y\in U_{i,y}\subseteq V_i$ such that $\pi_{i,y}\colon \pi^{-1}(U_{i,y})\to U_{i, y}$ satisfies $P$. But then $(U_{i,y})_{i\in I, y\in V_i}$ is an affine open cover of $Y$ such that the restrictions $\pi_{i,y}$ satisfy $P$, So by (ii) implies (i), $\pi$ satisfies $P$.
For a counterexample to (a), you can take the property $P$ to be "$Y$ can be covered by affine opens, each isomorphic to $\mathbb{A}^1$." It's easy to see that this property is local on the base. But it does not satisfy condition (a) of local on the target. Indeed, the identity map $\mathbb{A}^1\to\mathbb{A}^1$ satisfies $P$, but letting $V$ be the open subset of $\mathbb{A}^1$ obtained by removing a point, the restriction $V\to V$ does not satisfy $P$, since $V$ doesn't have any open subsets isomorphic to $\mathbb{A}^1$.
This property $P$ seems pretty pathological to me (and not really "local"), which makes me feel like Vakil's definition is "better" for ruling it out. But I'm not an algebraic geometer, so I can't comment on whether the distinction between these definitions is relevant or just an oversight.
