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About the coherent topology wiki page

  1. Feb 11, 2009 #1


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    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 [itex]X_i\cap X_j[/itex] by X_i and X_j are the same, then the inclusion maps [itex]\iota_i:X_i\rightarrow \cup_iX_i[/itex] 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?
    Last edited: Feb 11, 2009
  2. jcsd
  3. Nov 10, 2010 #2
    It's not true; in the case of CW complexes, you need an additional condition that [itex]X_i \cap X_j[/itex] is closed in [itex]X_i[/itex] for each i, j. Counterexample (due to my officemate Scott Van Thuong):

    Take a triangle [itex]\Delta ABC[/itex], and let [itex]X_1 = \overline{AB}[/itex], [itex]X_2 = \overline{BC}[/itex], and [itex]X_3 = \overline{CA}[/itex]. Give [itex]X_1[/itex] the Euclidean topology, and give both [itex]X_2[/itex] and [itex]X_3[/itex] the indiscrete topology. Let [itex]X[/itex] denote the whole triangle with the coherent topology. Now take an open neighborhood [itex]U[/itex] in [itex]X_1[/itex] with [itex]A \in U[/itex] but [itex]B \not \in U[/itex]. If the inclusion [itex]X_1 \hookrightarrow X[/itex] were an embedding, there would exist some open set [itex]V \subseteq X[/itex] with [itex]U = V \cap X_1[/itex]. Since V contains B, it must contain all of [itex]X_2[/itex]; therefore [itex]C \in V[/itex]. Since V contains C, it must contain all of [itex]X_3[/itex]; in particular, it contains A. But this contradicts the assumption.
    Last edited: Nov 10, 2010
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