# Finite non-Hausdorff models of CW-complexes

Years ago, my advisor showed me a construction where you take a CW complex and quotient each open cell to a single point. He said that under certain conditions (I believe always satisfied by the second barycentric subdivision of a complex) the resulting space would be a finite non-Hausdorff space homotopy equivalent to the original space.

Edit By 'homotopy equivalent' I really mean they have identical homotopy groups. I don't think this is equivalent to regular homotopy equivalence in the non-Hausdorff case.

He said this was well-known although not very popular. What is the name for this construction? What is a reference for the homotopy equivalence result?

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One name to google for is that of Jonathan Barmak. In particular, he has a Springer LNM on the subject which should be very useful. – Mariano Suárez-Alvarez Jul 17 '13 at 20:08
Maybe this link will be helpful too: ncatlab.org/nlab/show/finite+topological+space – Mathieu Baillif Jul 17 '13 at 20:08
What happened to your advisor and why cannot you ask him/her? – Mark Sapir Jul 17 '13 at 20:44
Jim Cannon retired and is composing music and writing a book. He answers e-mails but seems to be enjoying his retirement. He retired to get more time away from school duties and work on things he's behind on. – Brian Rushton Jul 17 '13 at 20:52
Here is an answer by Tom Goodwillie which gives an argument using van Kampen/excision: mathoverflow.net/questions/28380/… – Ricardo Andrade Jul 17 '13 at 21:03

When $K$ is a finite simplicial complex, this was done by Michael C. McCord ''Singular homology groups and homotopy groups of finite topological spaces'', Duke Math. J. 33 (1966), 465-474. If $X_K$ denotes the finite space you described, McCord proves that the quotient map $q:K \to X_K$ is a quasifibration in the sense of Dold and Thom, with contractible point-inverses. Thus $q$ is a weak homotopy equivalence and induces isomorphisms in singular homology.