Existence of well defined map $R/J\to R/I$ implies $J\cong M\subseteq I$?

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I'm currently trying to prove by myself some proposition related to Hopf-Galois theory (from the paper "Galois Correspondences for Hopf Bigalois Extensions", by Peter Schauenburg). The details of this proposition are not important here, I just want to ask about some detail I need to clarify to finish my proof. I want to know if the following statement is true or not:

Let $R$ a ring. Let $I,J$ be ideals of $R$. If there's a well defined map $R/J\to R/I$, then $R/I$ is a quotient of $R/J$ (or is a quotient of some $M\cong R/J$) (hence $J\cong M\subseteq I$).

Is this true? This seems to hold when I think about it, but don't find a way to prove it rigurously. How can I prove if it's correct? Any help or hint will be appreciated, thanks in advance.

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Let $R=\Bbb Z\times\Bbb Z$, $I=((1,0))$, and $J=((0,1))$. Then $R/I$, $R/J$ and $\Bbb Z$ are all isomorphic, but $J\not\subseteq I$.

For another example take $R=\Bbb Z\times\Bbb Z\times\Bbb Z$, $I=\Bbb Z\times\{0\}\times\{0\}$, $J=\{0\}\times\Bbb Z\times\Bbb Z$. Here $R/I\cong \Bbb Z\times\Bbb Z$ and $R/J\cong\Bbb Z$. There is certainly a well-defined map $R/J\to R/I$ given by $1\mapsto (1,1)$, but $\Bbb Z\times\Bbb Z$ is not a quotient of $\Bbb Z$. (If the last fact is not clear, replace $\Bbb Z$ with a finite ring.)