Let $R$ be a noncommutative ring with unit, let $P$ be a projective left $R$-module, and denote $^{\vee}\!P := \,\_R\mathrm{Hom}(P,R)$. One often sees it written that projectivity implies an isomorphism
$$
\mathrm{ev}:\,^{\vee}\!P \otimes\_R P \to R, ~~ \phi \otimes p \mapsto \phi(p).
$$
But I don't see that this is well defined! Consider
$$
(\phi.r) \otimes p \mapsto \phi.r(p) = (\phi(p))r
$$
where we have used the definition of the right $R$-module structure of $^{\vee}\!P$. Compare this with 
$$
\phi \otimes (r.p) \mapsto \phi(r.p) = r \phi(p),
$$
where we have used the fact that $\phi$ is a left $R$-module map. Now since $R$ is not commutative, these two values are not guaranteed to be equal, and the evaluation map is not guaranteed to be well-defined. What is wrong here?



Dual of a projective module
You are right. There is no such a map as the one you are trying to describe.


Here is a map that actually exists. Let $R$ and $S$ be two noncommutative rings with units, and let $P$ be an $R$-$S$-bimodule. Consider the $S$-$R$-bimodule $Q={}\_R\mathrm{Hom}(P,R)$. Then the evaluation is an $R$-$R$-bimodule map
$$
 \mathrm{ev}\colon P\otimes\_S Q\to R, \qquad (p\otimes\phi)\mapsto \phi(p),
$$
with the tensor product taken over the ring $S$.


In particular, if you do not have an $R$-$S$-bimodule but only a left $R$-module $P$, you can take $S=\mathbb Z$. Then you get an $R$-$R$-bimodule map
$$
 \mathrm{ev}\colon P\otimes\_{\mathbb Z}Q \to R,
$$
with the tensor product taken over the ring integers.


Notice that the $S$-$R$-bimodule $Q$ is (generally speaking) very different from the $S$-$R$-bimodule $Q'=\mathrm{Hom}\_S(P,S)$. For the bimodule $Q'$, the evaluation is an $S$-$S$-bimodule map
$$
 \mathrm{ev}\colon Q'\otimes\_R P\to S, \qquad (\psi\otimes p)\mapsto \psi(p).
$$


