why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I$?

Question

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 $\varphi \in \operatorname{End(R)}$ 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 $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

My question is:

Here is the solution:

1- I do not understand why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I, $ I can not see the details of this proof, could anyone explain that for me please?

2- I do not understand the definition of $\operatorname{Aut(R/I)}$ used in showing that $\bar{varphi} \in \operatorname{Aut(R/I)},$ Could anyone explain that for me please? I know that $\operatorname{Aut(R/I)}$ is the set of all isomorphisms from $R/I$ to $R/I$ but how is that definition related to the existence of inverse for $\bar{varphi},$ could anyone explains that for me please?

Answer 1

You wrote that we have $\varphi^{-1} \in \operatorname{Aut}_I(R)$. By definition, this means that $\varphi^{-1}(I)\subset I$, because this is meant by the subindex $I$: " Let $\operatorname{End_{I}(R)}$ be the set of $\psi \in \operatorname{End(R)}$ such that $\psi(I) \subset I.$ " So take $\psi=\varphi^{-1}$. So the claim from the title follows.