Talk:Hurewicz theorem

I thought I ought to update this article because the Hurewicz Theorem is a basic fact in algebraic topology, and both the absolute and relative forms are important. The proofs in G.W. Whitehead's book are fairly complicated, involving an induction via the Homotopy Addition Theorem for a simplex.

That theorem really involves the algebra of crossed complexes, and this is made explicit in the generalisation of the relative theorem given in the cited paper by Brown and Higgins.

This generalisation gave a clue to the joint work with Loday, and where the triadic form of the theorem has currently no other proof than via a (more general) Higher Homotopy van Kampen theorem.

Ronnie 20:54, 27 July 2007 (UTC)[reply]

There is a mistake in the statement of the the Absolute Hurewicz Theorem. It says that if X is an (n-1)-connected space, then the Hurewicz homomorphism is an epimorphism from the (n+1)th homotopy group onto the (n+1)th homology group. A counterexample, though, is the torus T. T is certainly path-connected (0-connected), and the second homotopy group of T is the trivial group, but the second homology group of T is the group of integers Z. So the Hurewicz homomorphism cannot possibly be an epimorphism.

For the moment, I am removing this part of the theorem statement (which does not seem to be included in any standard textbook treatments such as Spanier). If anyone knows what was intended, please upload a corrected statement. Cgd8d (talk) 01:13, 1 May 2008 (UTC)[reply]

This second part of Hurewicz (surjectivity) is only true when n > 1.

I think that the absolute theorem as stated is incorrect, however feel free to correct me if im wrong, since I am a number theorist and not a topologist. Let X be a (0)-connected space (i.e. a path-connected space). Then the theorem as stated says that \pi_0(X) is isomorphic to H_0(X), and \pi_1(X) is isomorphic to H_1(X). However, X is path-connected so that \pi_0(X)=0 and H_0(X)=Z (the integers), and these are certainly not isomorphic. Also if \pi_1(X) is non-abelian (for example X= figure of eight), then it is certainly not isomorphic to H_1(X), which is abelian.

To fix this problem I think that you need to use reduced homology and stipulate that n > 1. —Preceding unsigned comment added by 122.172.8.76 (talk) 11:53, 2 January 2010 (UTC)[reply]