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 \begin{align} &(1) A\ is\ flat\\ &(2) For\ every\ v \in K, there\ is\ an\ R-maps\ \theta:F\rightarrow K \ with\ \theta(v)=v. \end{align}

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?

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?