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:

$A$ is flat

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?

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environment (meant for math mode) to achieve the effect. $\endgroup$ – LSpice Nov 27 '18 at 18:53single$\theta$ that works forall$v$; but if every $v$ requires different $\theta$ and you cannot find one $\theta$ that would work for all of them, then you cannot conclude that $A$ is a direct summand. $\endgroup$ – მამუკა ჯიბლაძე Dec 5 '18 at 9:09