Contraction of Extension of Contraction of Ideal is Contraction

Theorem

Let $f : A \to B$ be a ring homomorphism.

Let $\mathfrak b$ be an ideal of $B$.

Let ${\mathfrak b}^c$ be the contraction of $\mathfrak b$ by $f$.

Let ${\mathfrak b}^{ce}$ be the extension of ${\mathfrak b}^c$ by $f$.

Let ${\mathfrak b}^{cec}$ be the contraction of ${\mathfrak b}^{ce}$ by $f$.

Then:

${\mathfrak b}^c = {\mathfrak b}^{cec}$

Since $\mathfrak b \supseteq {\mathfrak b}^{ce}$ by Ideal Contains Extension of Contraction:

$\ds {\mathfrak b}^c$

$\supseteq$

$\ds \paren { {\mathfrak b}^{ce} }^c$

$\ds $

$=$

$\ds {\mathfrak b}^{cec}$

$\ds $

$=$

$\ds \paren { {\mathfrak b}^c}^{ec}$

$\ds $

$\supseteq$

$\ds {\mathfrak b}^c$

$\blacksquare$