About the coherent topology wiki page

[quasar987]

On the wiki page on coherent topology, and more precisely, topological union (aka topology generated by a collection of spaces) (http://en.wikipedia.org/wiki/Coherent_topology#Topological_union), it is said that if the generating spaces {X_i} satisfy the compatibility condition that for each i,j, the subspace topologies induced on X_i\cap X_j by X_i and X_j are the same, then the inclusion maps \iota_i:X_i\rightarrow \cup_iX_i are topological embeddings (i.e. homeomorphisms onto their images).

Is this true? I tried proving it but without success but could not find a counter example either.

For instance, CW-complexes are the topological union of their n-skeletons. Is it always true that the inclusion of a n-skeleton into the CW-complex is a topological embedding?

 

[dan131m]

It's not true; in the case of CW complexes, you need an additional condition that X_i \cap X_j is closed in X_i for each i, j. Counterexample (due to my officemate Scott Van Thuong):

Take a triangle \Delta ABC, and let X_1 = \overline{AB}, X_2 = \overline{BC}, and X_3 = \overline{CA}. Give X_1 the Euclidean topology, and give both X_2 and X_3 the indiscrete topology. Let X denote the whole triangle with the coherent topology. Now take an open neighborhood U in X_1 with A \in U but B \not \in U. If the inclusion X_1 \hookrightarrow X were an embedding, there would exist some open set V \subseteq X with U = V \cap X_1. Since V contains B, it must contain all of X_2; therefore C \in V. Since V contains C, it must contain all of X_3; in particular, it contains A. But this contradicts the assumption.