Let $A$ be a noncommutative algebra over a field, say $\mathbb{C}$ or $\mathbb{R}$, and let $L$ be a bimodule over $A$. If $L$ is **invertible**, that is, if the dual right $A$-module $L^\*$ satisfies
$$
L^\* \otimes\_A L \simeq A
$$
then will $L$ necessarily be projective a left module over $A$? 



Invertible bimodules and projectivity
Yes. If $\varphi :L^\*\otimes \_A L\rightarrow A$ is an isomorphism, there exists an element $t =\sum\limits\_{i=1}^{n} x\_i^\*\otimes x\_i$ of $L^\*\otimes \_A L$ such that $\varphi (t )=1$. For any $x\in L$, we have then $x\varphi (t)=x$, that is, $\sum\_i \langle x,x\_i^\*\rangle x\_i=x$. Now consider the homomorphisms $u:A^n\rightarrow L$ and $v:L\rightarrow A^n$ defined by $u(e\_i)=x\_i$ and $v(x)=(\langle x,x\_1^\*\rangle,\ldots ,\langle x,x\_n^\*\rangle)$; we have $u\circ v=\operatorname{Id}\_L $, thus $L$ is a direct summand of $A^n$, hence projective.


