# A new unexpected result (ERROR!)

The below is wrong! The proof requires $\langle g^{-1}\rangle J$ to be a principal filter what does not necessarily hold.

I knew that composition of two complete funcoids is complete. But now I’ve found that for $g\circ f$ to be complete it’s enough $f$ to be complete.

The proof which I missed for years is rather trivial:

$\bigsqcup S \mathrel{[g \circ f]} J \Leftrightarrow J \mathrel{[f^{- 1} \circ g^{- 1}]} \bigsqcup S \Leftrightarrow \langle g^{- 1} \rangle J \mathrel{[f^{- 1}]} \bigsqcup S \Leftrightarrow \bigsqcup S \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[f]} \langle g^{- 1} \rangle J \Leftrightarrow \exists \mathcal{I} \in S : \mathcal{I} \mathrel{[g \circ f]} J$

Thus $g\circ f$ is complete.

I will amend my book when (sic!) it will be complete.