How do we understand the left and right module structure of Endomorphism ring I am reading some materials about noncommuative algebra which made me very confused.
Let $f: A\to B$ be a injective ring homomorphism between two rings, $A$, $B$ are not necessarily commutative. Then the author writes that:

Let $r:B\to E=\mathrm{End}_{A}(B)$ denote the canonical inclusion $b\to r_{b}$.

I do not understand the sentence. Since we have map $i$, we could realize $B$ as a two side module of $A$. What is the meaning of $\mathrm{End}_{A}(B)$ here? It means left $A$-module homomorphism or right $A$-module homomorphism or two side $A$ homomorphism?. Also what is $r_{b}$? Does it means $r_{b}(b’)=bb’$ or $r_{b}(b’)=b’b$? I am very confused. Can help me? Thanks a lot. By the way, could you recommend some references with details about this? The book which I am reading does not give any details.

Typically, if not specified, "module" is taken to mean "left module".  Indeed, sometimes people don't even work with right modules at all and instead think of a right module over a ring $A$ as a left module over the ring $A^{op}$ (which is $A$ with the order of its multiplication reversed).  This gives a unified way of treating both left and right modules together so that you don't need to worry about confusing them notationally.
So, in your quote, $B$ is presumably being considered as a left $A$-module, so $\operatorname{End}_A(B)$ refers to the left $A$-module homomorphisms $B\to B$.  This is confirmed by the choice of notation $r_b$: here $r$ stands for "right multiplication", so $r_b$ is the map that takes an element and multiplies it on the right by $b$.  That is, $r_b(b')=b'b$.  It is common to use $r$ or $\rho$ to denote right multiplication in this way while left multiplication would instead be denoted $l$ or $\lambda$.  Right multiplication commutes with left multiplication (this is just what associativity says: first multiplying by something on the right and then multiplying by something else on the left is the same as doing it in the opposite order), so right multiplication gives a homomorphism of left modules.  Explicitly, $r_b$ is a left $A$-module homomorphism since if $a\in A$ then $r_b(ab)=(ab)b'=a(bb')=ar_b(b')$.
