# Ideal:Kernel :: Filter:?

I understand that the notion of a filter is in some sense dual to the notion of an ideal, at least in the context of Boolean algebras1.

Let $f:{\mathbf A} \to {\mathbf B}$ be a Boolean algebra homomorphism. It's easy to verify that the the set $${\mathbf I} = f^{-1}(\{0\})$$ is an ideal in ${\mathbf A}$. The set ${\mathbf I}$ thus defined is commonly known as the kernel of $f$.

Similarly, it is easy to verify that the set $${\mathbf F} = f^{-1}(\{1\})$$ is a filter in ${\mathbf A}$.

My question is: Is there a standard name (a counterpart/dual of "kernel") for the set ${\mathbf F}$ thus defined?

(I've seen the term cokernel in other contexts, but I believe in this case the cokernel of $f$ would be the quotient ${\mathbf B}/\text{Im}(f)$, and it's not even obvious to me that such quotient would exist, let alone that an isomorphism between it and ${\mathbf F}$ would exist.)

1 An ideal in a Boolean algebra ${\mathbf A}$ is defined as a subset ${\mathbf I} \subseteq {\mathbf A}$ such that

1. $0 \in {\mathbf I}$;
2. if $p\in {\mathbf I}$ and $q\in {\mathbf I}$, then $p\vee q \in {\mathbf I}$;
3. if $p\in {\mathbf I}$ and $q\in {\mathbf A}$, then $p\wedge q \in {\mathbf I}$.

The definition of a filter in ${\mathbf A}$ is just the dual of the definition above. Namely, a filter is defined as a subset ${\mathbf F} \subseteq {\mathbf A}$ such that

1. $1 \in {\mathbf F}$;
2. if $p\in {\mathbf F}$ and $q\in {\mathbf F}$, then $p\wedge q \in {\mathbf F}$;
3. if $p\in {\mathbf F}$ and $q\in {\mathbf A}$, then $p\vee q \in {\mathbf F}$.
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Surely not standard in module theory, where the cokernel of a morphism $f\colon X\to Y$ is the quotient $Y/f(X)$. – egreg Sep 18 '13 at 17:32