$A$ is an associative algebra and $V$ is a representation. Then $\operatorname{End}_{A}(A)=A^{op}?$ (exercise from notes by Etingof)

Question

Let $A$ be an associative algebra. If $V$ is a representation of $A$, write $\operatorname{End}_A(V)$ to denotes the algebra of all homomorphisms of representations $V \to V$ . Show that $\operatorname{End}_A(A) = A^{op}$,the algebra A with opposite multiplication.
If $A$ is an associative algebra Show that $End_{A}(A)=A^{op}$ the algebra with opposite multiplication

In this link,there is an answer: fix $a∈A$ and denote by $R_a:x↦xa$ the right multiplication by $a$ . Then $\operatorname{End}_A(A)=\{R_a∣a∈A\}$. Show that this is isomorphic to $A^{op}$.
But I think $R$ is not a representation because $R_a R_b=R_{ba} ≠ R_{ab}$. So who is right?

Answer 1

$A^{op}$ has the same underlying set as $A$,with multplication given by: $$ a\times_{op}b:=ba $$ Now the map $;$ \begin{align} \psi:&A^{op}\to End(A) \\ \psi(a)&= R_{a} \end{align} satisfies : $$ \psi(a\times_{op}b)=\psi(ba)=R_{a}R_{b} $$ i.e $\psi$ gives an algebra isomorphism between $A^{op}$ and $End(A)$.Another equivalent approach is to consider the "evaluation" map: \begin{align} ev:&End(A) \to A^{op} \\ ev(T)&=T(1) \end{align}