# Final maps with the same domain

Let $$\mathbb D$$ and $$\mathbb E$$ be two directed sets, then a map $$f:\mathbb D\to \mathbb E$$ is said to be final,if for any $$e\in E$$ there exists some $$d\in D$$ such that $$f(d')\geq e$$ whenever $$d'\geq d.$$

Question: For two arbitrary directed sets $$\mathbb D$$ and $$\mathbb E,$$ does there exists a third directed set $$\mathbb A$$ such that there exists two final maps $$f_1:\mathbb A\to \mathbb D$$ and $$f_2:\mathbb A\to \mathbb E?$$ Any help would be appreciated.

• It seems like this third directed set would act like the categorical product of the first two.. is that true ;-) – Musa Al-hassy Dec 24 '18 at 15:50

Let $$\mathbb{A}$$ be the family of finite subsets of $$\mathbb{D}\cup \mathbb{E}$$ ordered by inclusion. This is clearly a directed set. Let $$f_1:\mathbb{A}\rightarrow \mathbb{D}$$ be a function such that, for every finite set $$F\in\mathbb{A}$$, $$f_1(F)\geq x$$ for every $$x\in F \cap \mathbb{D}$$. This can be done using definable choice because $$\mathbb{D}$$ is directed.

Try to define $$f_2$$ analogously and prove that this works.

• Excellent! BTW, I think it would be better to define $\mathbb A:=\left\{X\cup Y|\ X\ \text{is a finite subset of } \mathbb D,\ Y\ \text{is a finite subset of } \mathbb E\right\},$ in order to avoid the case $F\cap \mathbb D=\emptyset$ or $F\cap\mathbb E=\emptyset.$ – painday Dec 25 '18 at 4:37
• For my supervisor zero is still a finite cardinality. Anyhow the definition I gave still works if $F \cap \mathbb{D}$ is empty. In that case $f_1(F)$ would just be any element in $\mathbb{D}$. – Anguepa Dec 26 '18 at 17:55