$\operatorname{Hom}_A(A^*, P)\cong P$ such that $P$ is projective under the given assumption?

Question

Let $A$ be a finite dimensional algebra over a field $k$. Let $A^* = \operatorname{Hom}_k (A,k)$ and $A^*$ is $A$-bimodule. Suppose that $\operatorname{Hom}_A (A^*, A) \cong A$ as left $A$-module. Then is it true that $\operatorname{Hom}_A(A^*, P)\cong P$ such that $P$ is projective?

Answer 1

Since $A^*$ is finite dimensional as a $k$-vector space, then $A^*$ is finitely generated both as a left and as a right $A$-module.

In particular we can state that $$ \textstyle\operatorname{Hom}_A(A^*,\bigoplus\limits_{i\in I}M_i)\cong \bigoplus\limits_{i\in I}\operatorname{Hom}_A(A^*,M_i) $$ for any index set $I$ and any family of $A$-modules. If we denote by $A^{(I)}$ a direct sum of copies of $A$, we get $$ \operatorname{Hom}_A(A^*,A^{(I)})\cong\bigl(\operatorname{Hom}_A(A^*,A)\bigr){}^{(I)} $$ so your statement is true for every free $A$-module. If $P$ is projective, then $P\oplus P'$ is free, for some module $P'$.