In the textbook "An introduction to homological algebra" written by Rotman, there is an exercise 3.12 in p.115:
If $P$ is a finitely generated left $R$-module, prove that $P$ is projective if and only if $\operatorname{id}_P\in\operatorname{im}\nu$, where $\nu:\operatorname{Hom}_R(P,R)\otimes_R P \rightarrow \operatorname{Hom}_R(P,P)$ is defined, for all $x\in P$, by $f\otimes x\mapsto \tilde{f}$, where $\tilde{f}:y \mapsto f(y)x$.
However, I don't know how to find $f$ and $x$ satisfying $f(y)x=y$ for all $y\in P$. Help me.
Let $n\in\Bbb N$, $\varrho:R^{\oplus n}\to P$ be a surjective $R$-module homomorphism, $\sigma:P\to R^{\oplus n}$ be a section, $\pi_i:R^{\oplus n}\to R$ for $i<n$ be the canonical projection and $e_i\in R^{\oplus n}$ for $i<n$ be the canonical base. For every $y\in P$ we have $$\sigma(y)=\sum_{i<n}(\pi_i\circ\sigma)(y)e_i$$ Let $$z=\sum_{i<n}(\pi_i\circ\sigma)\otimes_R\varrho(e_i)\in\operatorname{Hom}_R(P,R)\otimes_RP$$ Then we have \begin{align} \nu(z)(y) &=\sum_{i<n}(\pi_i\circ\sigma)(y)\varrho(e_i)\\ &=\varrho\left(\sum_{i<n}(\pi_i\circ\sigma)(y)e_i\right)\\ &=(\varrho\circ\sigma)(y)\\ &=y \end{align}