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Say I have some function $f:R \rightarrow S$ such that $f$ is a ring homomorphism and $J$ is an ideal of $f$. $I = f^{-1}(J)$ is an ideal of $R$, but I don't really understand why. $J$ being an ideal means that $I$ is non-empty, that much makes sense.

But how can I show that addition/absorption hold if $f$ isn't surjective, nor do I know if $f^{-1}$ is even a ring homomorphism? I've seen a very similar question has been asked, but the only response appeared to just prove that these things held in $J$.

For instance, if I want to prove addition holds, I could take $i_1, i_2 \in I$ so that for $j_1, j_2 \in J$, $i_1 = f^{-1}(j_1)$, $i_2 = f^{-1}(j_2)$. But then I can't really add these elements together because I don't know if addition is preserved in $f^{-1}$. What gives?

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The definition of $f^{-1}(J)$ is $\{r\in R: f(r) \in J\}$. If $i_1,i_2 \in f^{-1}(J)$ the $f(i_1)$ and $f(i_2) \in J$ so $f(i_1+i_2)=f(i_1)+f(i_2) \in J$ which implies $i_1+i_2 \in f^{-1} (J)$. It is not assumed that $f$ is a bijection so you cannot think of $f^{-1}$ as a function from $S$ into $R$. You have to use definition of inverse image of a set: $f^{-1}(J)=\{r\in R: f(r) \in J\}$. Now you should be able to complete the rest of the proof.

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Suppose $x,y\in f^{-1}(J)$.

Then $f(x),f(y)\in J$ so $f(x)-f(y)=f(x-y)\in J$.

So $x-y\in f^{-1}(J)$.

If $a\in R, x\in f^{-1}(J)$.

Then $f(ax)=f(a)f(x) \in J$, since $f(a)\in S$ and $f(x)\in J$.

So $ax\in f^{-1}(J)$, and similarly for the reverse multiplication

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