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if $a$ is a nonzero element of projective $R-module$ $P$ ; prove that:

homorphism $f$ $:$ $P$ $\to$ $R$ is exist where $f(a)$ $\neq$ $0$.

my work :
proof by contradiction ; suppose $f(a)$ $=$ $0$
since $P$ is projective $R-module$ then exist homorphism $h$ $:$ $P$ $\to$ $A$ where if $g$ $:$ $A$ $\to$ $R$ is arbitrary homomorphism then $gh$$=$$f$
can you help me for I continued my work ?

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    Maybe you should consider the map $R \to P, \lambda \to \lambda a$. – 2017-01-04
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    One way of seeing this is to prove for free R-modules first. – 2017-01-04

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$P$ projective means there is a module $P'$ such that $P\times P'\cong R^n$ for some natural $n$. Let $\phi$ be this isomorphism. Then define $f':P\to R^n$ by $f'(p)=\phi(p,0)$. Because $a \neq 0$ and $f'$ is injective, $f'(a) \neq 0$, and therefore, at least one of the components of $f'(a)$ must be non-zero. Let $\pi_i:R^n\to R$ be projection into this component. Then $f=\pi_i\circ f'$ is the homomorphism you're after.