A module $P$ is projective if and only if for every diagram 
there exists a homomorphism $\tilde h : P \to N$ that lifts $h$, i.e., such that $g \circ \tilde h = h$
I know many sources give this condition know as the lifting property as the definition of projective functor but mine is
Def: $R$-module $P$ is defined to be projective if the functor Hom$_R(P, −)$ : R-Mod $\to$ Ab is exact.
I have already one way. Now I am stuck in the proving that if Hom$_R(P, −)$ is exact, that is if I have a projective module according to my definition, the lifting property holds. Could you show me how to prove it or point me to a proof?
That Hom$_R(P,-)$ is exact means is that for every s.e.s $0\to N\to N' \to N''\to 0$, the exact sequence $0\to Hom_R(P,N)\to Hom_R(P,N') \to Hom_R(P,N'')\to 0$ is exact. How do I continue from here?
We are given a surjection $g\colon N\to N'$ and a map $h\colon P\to N'$, and need to produce a map $\widetilde{h}\colon P\to N$ such that $h=g\widetilde{h}$. We extend the surjection $g\colon N\to N'$ to a short exact sequence $0\to \mathrm{ker}(g)\to N\xrightarrow{g} N'\to 0$. Since $\mathrm{Hom}_R(P,-)$ is assumed to be exact, we obtain another short exact sequence $0\to\mathrm{Hom}_R(P,\mathrm{ker}(g))\to\mathrm{Hom}_R(P,N)\xrightarrow{g\circ-}\mathrm{Hom}_R(P,N')\to 0$. In particular, the map $g\circ -\colon \mathrm{Hom}_R(P,N)\to\mathrm{Hom}_R(P,N')$ is surjective, so we can find a preimage of the element $h\in\mathrm{Hom}_R(P,N')$, which is an element $\widetilde{h}\in\mathrm{Hom}_R(P,N)$ such that $g\widetilde{h}=h$.