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This is a statement made in Eisenbud Commutative Algebra Chapter 2, localization Section 1.

If $\phi:R\to S$ is any map of sets, then operation taking subs of $S$ to subsets of $R$ by $I\to \phi^{-1}(I)$ preserves inclusion and intersections. If $\phi$ is a map of rings, and $I\subset S$ is an ideal. Then he concludes $\phi$ induces an injection $R/\phi^{-1}(I)\subset S/I$.

I am not clear about this injection here. I guess he means injection in the sense of ideals rather than elements.

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Consider the canonical projection $\pi: S \rightarrow S/I$. The kernel of $\pi \circ \phi$ is $\phi^{-1}(I)$, so by the homomorphism theorem, there is an induced injection $R/\phi^{-1}(I) \rightarrow S/I$.

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    Is this injection the injection of elementwise of the ring as homomorphism or injection as ideals of ring? That is my question. I can see this as an injection as the set of ideal. – 2017-02-03
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    this is an injective ring homomorphism. To see that this cannot be an injection on the set of ideals (I'm not sure if I understand you're question correctly), consider the case where $R = \mathbb{Z}$, $S = \mathbb{Q}$, $I = (0)$ and $\phi$ the standard embedding. Then $R/\phi^{-1}(I)=\mathbb{Z}$, $S/I=\mathbb{Q}$, and there are infinitely many ideals in $\mathbb{Z}$, but only two in $\mathbb{Q}$. Also $\mathbb{Z}$ is not an ideal of $\mathbb{Q}$, if that's your question. – 2017-02-03