Here is the question I want to prove:
Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\phi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$
$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\phi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$
$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$
My question is:
1- For letter $(a)$I know that if I used the first isomorphism theorem, I would get $\bar{\varphi} \pi = \varphi.$ But how can I get the result that I want above? could anyone help me in figuring this out please?
2- For letter $(b),$ I do not know exactly what I am proving, I feel like I am proving well defined of taking the bar operation. But what does this operation means? How can I prove the required?
Any help would be appreciated.
The definition of the "bar" operator is given in the first part of the problem. The only possibility for $\bar{\varphi}$ is $\bar{\varphi}(\pi(x))=\bar{\varphi}(x+I)=\varphi(x)+I = \pi(\varphi(x))$. Can you show that this is indeed an endomorphism of $R/I$ using the property that $\varphi(I)\subseteq I$?
For the second part you need to show that the bar you just defined respects the composition of endomorphisms. Perhaps now that you see the definition you can show that $\overline{\varphi\psi}=\bar{\varphi}\bar{\psi}$.