If 2 is a unit in a ring R, then there exists a bijection between the idempotents of the ring and the self-inverse elements

Question

Suppose $R$ is a ring with unity $1_R$, and suppose $2=1_R+1_R$ is a unit. I am asked to show that the map $\sigma: \{e:e^{2}=e\} \rightarrow \{u:u^{2}=1\},$ $\sigma(e)=1-2e$ is a bijection. Here is my attempt: First, the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $\sigma$ is indeed self-inverse. Injectivity is clear. To show surjectivity, suppose $u\in R$ with $u^{2}=1$. Consider the element $2^{-1}(1-u)$. $1-2(2^{-1}(1-u))=1-(1-u)=u$, so if we can show $2^{-1}(1-u)$ is an idempotent, we're done. To establish this, note first that since $u$ commutes with $2$ it must also commute with $2^{-1}$. Hence we may write $(2^{-1}(1-u))^{2}=(2^{-1})^{2}(1-u)^{2}=(2^{-1})^{2}(1-2u+u^{2})=(2^{-1})^{2}(2-2u)=2^{-1}(1-u)$, as desired. Is my proof correct? I welcome any comments or criticism.

Answer 1

the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $\sigma$ is indeed self-inverse.

Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."

To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $e\mapsto 1-2e$ defines a function since it is the composition of functions $e\mapsto 2e$ and $e\mapsto 1-e$."

That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.

Otherwise, I think you have covered all the bases well.