 Hom into the regular algebra as a right module

For a left module $M$ over a unital associative algebra $A$, right multiplication in $A$ gives a right module structure on ${\mathrm{Hom}}_A(M,A)$ by $(f\cdot a)(m)=f(m)a$. This remains a module homomorphism because $f(bm)a=bf(m)a$; associativity and the unit verify the action axioms. When $A=M_d(\mathbb{C})$ and $M={\mathbb{C}}^d$ is its simple module of columns, every such map is $m\mapsto mr$ for a unique row $r$, making the Hom space the simple row right module. This illustrates how left and right simple modules correspond for a semisimple algebra.