In Rotman's book "Intro to homological algebra" Theorem 3.62 
Let $0\rightarrow K\rightarrow\ F\rightarrow A\rightarrow 0$ be an exact sequence of right R-modules, where $F$ is free. The following are equivalent:


1. $A$ is flat
2. For every $v \in K$, there is an $R$-map $\theta:F\rightarrow K$ with $\theta(v)=v$.


My problem is that isn't the second condition implies that $A$ is a direct summand of free module $F$, hence projective. Then isn't this theorem is saying that flat implies projective? And we already know that projective modules are flat. So isn't this theorem saying that projective are equivalent to flat which is not true in general. For example $\mathbb Q$ as a $\mathbb Z$-module. So where am I wrong?



Flat module and Projective Module
The point is that the map $\theta$ is allowed to depend on $v$. So this is not a splitting of the exact sequence.


