Let $G$ be a finite group, $H\subset G$ a subgroup, $k$ a commutative ring, $M$ a $kG$-module, $n\in\mathbb{Z}$, and $\hat{H}\,^n(H,M)$ the $n$th Tate cohomology group as defined in this question, where we consider $M$ a $kH$-module via $kH\hookrightarrow kG$.
We say that $M$ is cohomologically trivial if $\hat{H}\,^n(H,M)=0$ for all $n\in\mathbb{Z}$ and all subgroups $H\subset G$. Using Shapiro's lemma and the double coset formula, I can see that free $kG$-modules are cohomologically trivial. I'd like to use this to conclude that projective $kG$-modules are also cohomologically trivial.
If $N$ is projective, then we can find $N'$ such that $N\oplus N'=M$ is free, and hence cohomologically trivial. Then, the injection of $kG$-modules $N\hookrightarrow M$ gives a map of cohomology groups
$$\hat{H}\,^n(H,N)\to\hat{H}\,^n(H,M)=0$$
but unless this map is injective, we can't conclude that $N$ is cohomologically trivial. Is this map injective, and if not, how can I show that projective modules are cohomologically trivial? Perhaps I need to use the fact that $N$ is a direct summand of $M$, and not just a submodule.
The inclusion of $N$ into $M$ has a section, that is, a map $M\to N$ whose composition with $N\to M$ is the identity of $N$. This and functoriality does what you want.