Let $A$ and $B$ be $R$-bimodules, and $\alpha$ an $R$-automorphism. Write ${}_1A_\alpha$ for the right-twisted $R$-bimodule with action $(r\cdot x\cdot s)\mapsto rx\alpha(s)$.

Is it true that a homomorphism $A\xrightarrow{\varphi}B$ is also a homomorphism ${}_1A_\alpha\xrightarrow{\varphi}{}_1B_\alpha$?

I'm asking this question becasue I do not have much time to find it for myself, and is really just a fact (if it's true) that I'd benefit from. I certainly think it is true, but it's all too easy for me to miss something, like a necessary constraint (or even a rather trivial counterexample that it isn't true in general). Is there a reference for this?


1 Answer 1


You certainly have figured this out. Just to lift this question from the unanswered questions pile.

Yes, this is true. Your twisted modules are an example of restriction of scalars, and this is functorial.

Since you're twisting just on the "right" structure, we might as well just think about right $R$-modules $A,B$, along with an $R$-automorphism $\alpha:R\to R$. Let $A_\alpha, B_\alpha$ be the restrictions of $A,B$ along $\alpha$, and $\varphi:A\to B$ an $R$-module homomorphism. Since $\varphi(a\cdot r)=\varphi(a)\cdot r$, for all $a\in A, r\in R$, we have $$ \varphi(a\cdot \alpha(r))=\varphi(a)\cdot \alpha(r)\,, $$ for all $a\in A, r\in R$. This makes the (restricted) map $\varphi: A_\alpha\to B_\alpha$ an $R$-module homomorphism as well.


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