About injectivity of induced homomorphisms on quotient rings Let $A, B$ be commutative rings with identity, let $f: A \rightarrow B$ be a ring homomorphism (with $f(1) = 1$), let $\mathfrak{a}$ be an ideal of $A$, $\mathfrak{b}$ an ideal of $B$ such that $f(\mathfrak{a}) \subseteq \mathfrak{b}$. Then there is a well-defined homomorphism
\begin{align}
&\bar f: &&A / \mathfrak{a} &&\rightarrow &&B / \mathfrak{b} \\
& &&a + \mathfrak{a} &&\mapsto &&f(a) + \mathfrak{b}.
\end{align}
It's clear to me that if $\mathfrak{a} = \mathfrak{b}^c$, then $f$ injective $\implies \bar f$ injective. 
Question: Under what conditions does $\bar f$ injective $\implies f$ injective hold?

Let $\bar f$ is injective. Then $f(a)\in \mathfrak{b} $ implies $a\in \mathfrak{a} $. This means $\mathfrak{a} \supseteq \mathfrak{b}^c$ hence $\mathfrak{a}=\mathfrak{b}^c$ .
