### A new easy proposition about funcoids

I have proved (see new version of my book) the following proposition. (It is basically a special case of my erroneous theorem which I proposed earlier.)

Proposition For $latex f \in \mathsf{FCD} (A, B)$, a finite set $latex X \in \mathscr{P} A$ and a function $latex t \in \mathscr{F} (B)^X$ there exists (obviously unique) $latex g \in \mathsf{FCD} (A, B)$ such that $latex \langle g\rangle p = \langle f \rangle p$ for $latex p \in \mathrm{atoms}^{\mathscr{F} (A)} \setminus \mathrm{atoms}\, X$ and $latex \langle g\rangle @\{ x \} = t (x)$ for $latex x \in X$.

This funcoid $latex g$ is determined by the formula

$latex g = (f \setminus (@X \times^{\mathsf{FCD}} \top)) \sqcup \bigsqcup_{x \in X} (@\{ x \} \times^{\mathsf{FCD}} t (x)) .$

and its corollary:

Corollary If $latex f \in \mathsf{FCD} (A, B)$, $latex x \in A$, and $latex \mathcal{Y} \in \mathscr{F} (B)$, then there exists an (obviously unique) $latex g \in \mathsf{FCD} (A, B)$ such that $latex \langle g\rangle p = \langle f \rangle p$ for all ultrafilters $latex p$ except of $latex p = @\{ x \}$ and $latex \langle g \rangle @\{ x \} = \mathcal{Y}$.

This funcoid $latex g$ is determined by the formula

$latex g = (f \setminus (@\{ x \} \times^{\mathsf{FCD}} \top)) \sqcup (\{ x \} \times^{\mathsf{FCD}} \mathcal{Y})$.