Some questions about standard $K$-duality

Question

Let $A$ be a finite dimensional $K$-algebra, where $K$ is an algebraically closed field.

We define a funtor $D: mod A \to mod A^{op}$ called standard $K$-duality. Suppose that $M$ is an arbitrary right $A$-module. Let $M^*=D(M)=Hom_{K}(M,K)$. I know that $M$ is isomorphic to $D(M)$ as vector spaces. I want to know if the following statements are right.

1. $A^{op}$ is isomorphic to $D(A)$ as left $A$-modules? If $A^{op} \cong D(A)$, how to define the isomorphism.
2. Suppose that $M \in mod A$, i.e. $M$ is finitely generated right $A$-moudle. If $M=(K \overset{1}{\rightarrow} K {\rightarrow} 0 $), then $D(M)=(K \overset{1}{\leftarrow} K {\leftarrow} 0)$. I have no idea why. In genarally, if $Q$ is a quiver, then $Q^{op}$ is a quiver which reverses all arrows in $Q$?

Answer 1

1. No. $A^{op}$, as a left $A$-module (that is to say, a right $A^{op}$-module), is projective, while $D(A)$ is injective. Algebras for which $A$ is isomorphic to $D(A)$ as a left module are called self-injective algebras.
2. For your first sentence, you seem to be implicitly assuming that $A$ is the path algebra of the quiver $1\to 2\to 3$. In any case, if $A$ is the path algebra of any quiver $Q$, then the opposite algebra $A^{op}$ is the path algebra of the opposite quiver $Q^{op}$ (obtained by reversing all arrows of $Q$). If $M$ is a representation of $Q$, then $D(M)$ will be a representation of $Q^{op}$, obtained by taking the dual vector space at each vertex and the dual linear map at each arrow.