# About the coherent topology wiki page

1. Feb 11, 2009

### 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?

Last edited: Feb 11, 2009
2. Nov 10, 2010

### 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.

Last edited: Nov 10, 2010