An easy property for extension and contraction of ideal

Question

So given ring homomorphism $f:A\rightarrow B,$ for commutative ring with identity. My problem is about the property $\mathfrak{a}\subseteq\mathfrak{a}^{ec}.$ It's easy to make myself convinced about this property since $\mathfrak{a}$ definitely lies in the preimage of ideal generated by $f(\mathfrak{a})$ (for $Bf(\mathfrak{a})$ simply contains $\mathfrak{a}$). But since I saw in Extension and Contraction of ideals the author mentioned $\mathfrak{a}^{ec}= \ker f\mathfrak{a} \supset \mathfrak{a}$ which makes me quite confused. Because since $\ker f$ and $\mathfrak{a}$ are two ideals so the elements in $kerf\mathfrak{a}$ are just formal linear sum $\sum x_ia_i$ where $x_i\in \ker f,a_i\in \mathfrak{a}.$ But once we put any element of this form into the homomorphism $f$ we have a ZERO. So I don't know why he said this. I also want to look for the relationship between $\ker f$ and $\mathfrak{a}$ and my natural idea is to consider the isomorphism theorem. Namely $A/\ker f\cong Imf,$ so we look at the image of $\mathfrak{a}$ in this case and we know that $f(\mathfrak{a})$ should have a 1-1 correspondence with $\pi(\mathfrak{a})$ here I use $\pi$ to denote the natural homomorphism between $A$ and $A/\ker f.$ But my thoughts just stop here since I don't know how to move on, so if anyone know a connection please give me a hint. (I didn't find any useful information.)