Let $R$ be a not-necessarily commutative ring, and suppose $M$ is left $R$-module. Let $R^\text{op}$ consist of elements written as $a^\text{op}$, where as an element $a^\text{op} = a$ for all $a\in R$, but $a^\text{op} b^\text{op} = ba$. The ring $\text{End}_R(M)$ is the ring of $R$-module homomorphisms $M \to M$. Today, I heard the claim that $\Phi: R^\text{op} \to \text{End}_R(M)$ defined by $a^\text{op} \mapsto (x \mapsto ax)$ was a valid map. However, I have several questions:
- The map $\varphi_a : M \to M$ for $a\in R$ defined by $x \mapsto a\bullet x$ sure doesn't look $R$-linear, because $\varphi_a(b\bullet x) = a\bullet (b\bullet x)$ where "$\bullet$" denotes left multiplication of $R$ on $M$. On the other hand, $b \bullet \varphi_a(x) = b \bullet (a \bullet x)$. Moreover I don't think this is an issue that can be solved by taking $a^\text{op}$, because $M$ is a left $R$-module, not $R^\text{op}$-module. I know here For a ring $R$, does $\operatorname{End}_R(R)\cong R^{\mathrm{op}}$?, they considered $\varphi_a : M \to M$ defined by $x \mapsto x a$ (i.e. right multiplication), but that doesn't work here either because $M$ is a LEFT $R$-module.
- Not sure why we need $R^\text{op}$, because (taking $\Phi: R \to \text{End}_R(M)$ defined by $a \mapsto (x \mapsto ax)$ instead), we have that $\Phi(ab) = [x \mapsto (ab)\bullet x] = \Phi(a) \circ \Phi(b)$, which looks totally fine to me in terms of showing that $\Phi$ is a ring homomorphism. So I was confused because I feel like $R^\text{op}$ vs. $R$ is only dependent on the notation you choose for function composition (the "standard" right-to-left way or the "intuitive" left-to-right way). EDIT: for example, following the above link using right-multiplication, we have that $$\varphi_a(\varphi_b(x)) = \varphi_a(xb) = (xb)a = x (ba) = \varphi_{ba}(x),$$ and so we get $a \circ_{\text{End}_R(M)} b = ba$. In this notation, it is clear that we have opposite. But, if we use instead the notation $((x)\varphi_b)\varphi_a$, we shouldn't need the opposite notation, right?
As Geoffrey explained, in order for this to hold, $M$ would have to be an $(R,R)$-bimodule.
In general, if $M$ is a left $R$-module, then it is naturally a right $R^{op}$-module by the obvious action: $x\cdot a^{op}:= a\cdot x$.
$R$ itself is both a left $R$-module and a right $R$-module. One may consider the right $R$-module structure as a left $R^{op}$-module the same way as above and so one does have a natural ring homomorphism: $$ \phi: R^{op} \to End_R(R), $$ given for $a\in R$, $$ \phi(a)=(x\mapsto x\cdot a) \quad (\text{and note that } x\cdot a= a^{op}\cdot x). $$
So, why does this give an $R$-endomorphism? This is indeed, because $R$ is an $(R,R)$-bimodule: Further let $r\in R$ and $x\in R$. Then $$ \big(\phi(a)\big) (r\cdot x)= (r\cdot x) \cdot a = r\cdot (x\cdot a) = r\cdot \big(\phi(a)\big) (x). $$
And why $R^{op}$ and not $R$? It is because that is how this gives a ring homomorphism: $$ \big(\phi(a)\circ \phi(b)\big)(x) = \big(\phi(a)\big) \bigg(\big(\phi(b)\big)(x)\bigg)= \big(\phi(a)\big) \big(x\cdot b\big)= (x\cdot b) \cdot a= x\cdot (ba) =\big(\phi(ba)\big)(x). $$
So $\phi(b)\circ \phi(a)=\phi(ba)$, and hence this is a ring homomorphism only if $a\star b= ba$, i.e., if you are mapping from $R^{op}$ and not $R$!
Regarding the notation of endomorphisms: indeed if you use the second notation, then the suggestive multiplication gives you $\big(End_R(R)\big)^{op}$. The point of this is that then you can actually think of mutiplication by an element of $R$ giving an endomorphism and not have to work with $R^{op}$.
This extends in a natural way to free $R$-modules, so one has a natural morphism $$ R^{op}\to End_R(F), $$ where $F$ is a free $R$-module, but unfortunately not to general modules.
Another situation in which this works to some extent is if $R$ is an $A$-algebra, where $A$ is a commutative ring, e.g., if $R=A[G]$, a group algebra. In this case, the left multiplication by an element of $R$ gives an $A$-algebra endomorphism of $M$, i.e., if $M$ is a left $R$-module, then one has a natural ring homomorphism, $$ \lambda: R \to End_A(M), $$ given by left-multiplication by elements of $R$: $$ \lambda(a)(x):= ax. $$ As you observed, this is not an $R$-module endomorphism, but it is an $A$-module homomorphism, because by the definition of an $A$-algebra, the image of $A$ in $R$ (mapping $1\in A$ to $1\in R$) is in the center of $R$. (In case of a group algebra, $A\subseteq Z(A[G])$.)