@MattN. I'd suspect $\phi$ to be the natural map from $R$ to $R/I$, $$\phi(r)=r+I.$$

Then if $J$ is an ideal in $R/I,$ we also have that $\phi^{-1}(J)$ is an ideal in $R,$ such that $J\supseteq I$.

The bijection is between ideals in $R/I$ and ideals in $R$ containing $I$.

Hello ymar!

@ymar Yes : ) I think I figured it out this morning. It makes no difference whether he defines it to be $\pi : R \to R/I$ or $\pi^{-1}: R/I \to R$, right? He wrote $J \mapsto \pi^{-1}(J)$ but it would've been more obvious if he'd written $J \mapsto \pi(J) = J/I$.

In retrospect this is one of the stupidest questions I've asked.

Thank you! : )

@MattN Hey! Well, $\pi$ need not be an injection, so $\pi^{-1} :R/I\to R$ may not exist. What exists is an inverse image of an ideal $J\subseteq R/I$ under $\pi$, which we denote be $\pi^{-1}(J)$.

@ymar Well, $\pi : R \to R/I$ need not be injective. But if we view it as a map from ideals to ideals I think it becomes injective. (I proved it but my proof might be wrong).

Define $\pi: J \mapsto J/I$.

@Matt Sure it is! :)

Then I fail to see why the lecturer would write $J/I \mapsto \pi^{-1}(J/I)$ instead of $J \mapsto J/I$. Oh well... : )

One can always write things extra complicated so that simple stuff becomes extra confusing, I guess.

Because he wasn't thinking of $\pi$ as a map from a set of ideals to a set of ideals.

But he wrote:

$$ \{ \text{ ideals of A/I} \} \leftrightarrow \{ \text{ ideals of A containing I} \}$$

Then

$$ J \mapsto \phi^{-1}(J)$$

: /

Right! See, the map $\phi$ goes from $A$ to $A/I$. Saying it goes from $\{ \text{ ideals of A containing I} \}$ to $\{ \text{ ideals of A/I} \}$ would be imprecise. $\phi$ does induce such a function, but it is not the function. But writing $\phi^{-1}(J)$ is correct, because it means the inverse image of $J$ under $\phi$.

Ah.I see. I actually meant "it induces" : )

Thank you!

You're welcome. :)