# Posets – \$ X \$, \$ Y \$ spaces such that the cones on \$ X cap Y \$ generate \$ X cup Y = sum (X cap Y) \$

I have this question from Kratzer and Thévenaz – Type of motor surgery of the exercises and the activities of the groups of people:

On page 89, they explain that

Yes $$Y = Y_1 cup Y_2$$ They are such that $$X = Y_1 cap Y_2$$ and if each one $$Y_i$$ it's a cone about $$X$$, so $$Y$$ is the suspension of $$X$$

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

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