I am studying algebra from Thomas Hungerford and I was unable to understand reasoning behind this particular corollary (Corollary 6.3) Page 219.
Corollary 6.3. A unitary module $A$ over a principal ideal domain is free if and only if $A$ is projective.
Proof. ($\Rightarrow$) Theorem 3.2. ($\Leftarrow$) There is a short exact sequence $0\to K\xrightarrow{\subset}{F}\xrightarrow{f}{A}\to 0$ ith $F$ free, $f$ an epimorphism and $K=\operatorname{Ker} f$ by Corollary 2.2. If $A$ is projective, then $F\cong K\oplus A$ by Theorem 3.4. Therefore, $A$ is isomorphic to a submodule of $F$, whence $A$ is free by Theorem 6.1. $\blacksquare$
Corollary 2.2: Every (unitary) module A over a ring R (with identity) is the homomorphic image of a free R-module F. If A is finitely generated, then F may be chosen finitely generated.
Question: How does unitary module or projective module implies existence of short exact sequence given in image?
I have no help as this was not covered in class.
Any help would be really appreciated.