# posets – The spaces \$ X \$, \$ Y \$ such that the cones over \$ X cap Y \$ involve \$ X cup Y = sum (X cap Y) \$

I have this question from Kratzer and Thevenaz – Homopopy type of lattices and lattices of a finite group:

On page 89, they explain that

Yes $$Y = Y_1 cup Y_2$$ are such that $$X = Y_1 cap Y_2$$ and if everyone $$Y_i$$ is a cone on $$X$$then $$Y$$ is the suspension of $$X$$

$$def abs # 1 { lvert # 1 rvert}$$But on page 90, in the proof of Proposition 2.5they use that in a way that I do not understand. They prove that $$abs X simeq operatorname C abs E$$ and $$abs Y simeq operatorname C abs F$$but they are not cones $$abs {X cap Y}$$, so why can I continue proof of this proposal saying that this implies $$abs G simeq sum ( abs {X cap Y})$$?

My concept of "cone on $$abs {X cap Y}$$" is:
$$abs {X cap Y} times [0,1]/ (a, i) thicksim (b, 1)$$
with $$a, b in X cap Y$$. Do I misunderstand what is a cone? $$abs {X cap Y}$$?