# Is there a gap in the standard treatment of simplicial homology?

On MO, Daniel Moskovich has this to say about the Hauptvermutung:

The Hauptvermutung is so obvious that it gets taken for granted everywhere, and most of us learn algebraic topology without ever noticing this huge gap in its foundations (of the text-book standard simplicial approach). It is implicit every time one states that a homotopy invariant of a simplicial complex, such as simplicial homology, is in fact a homotopy invariant of a polyhedron.

I have to admit I find this statement mystifying. We recently set up the theory of simplicial homology in lecture and I do not see anywhere that the Hauptvermutung needs to be assumed to show that simplicial homology is a homotopy invariant. Doesn't this follow once you have simplicial approximation and you also know that simplicial maps which are homotopic induce chain-homotopic maps on simplicial chains?

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Well, I agree that this statement is mystifying, and look forwards to some explanation. Simplicial homology is a homotopy invariant as you say. I can imagine that someone, somewhere, has "proved" that the Euler characteristic is homotopy invariant by subdividing simplicial complexes though (I don't know about standard textbooks doing this). – George Lowther Mar 3 '11 at 1:48
Have you tried asking Daniel Moskovich directly? – Chris Eagle Mar 3 '11 at 11:13
@Chris: good point. I've sent him an e-mail asking him to respond here. – Qiaochu Yuan Mar 3 '11 at 16:57
I thought it relevant to post the following from Rotman. After proving the following: Let $X$ be a polyderon having triangulations $(K,h)$ and $(K',h')$. Then $H_q(K) \simeq H_q(K')$ for all $q \ge 0$, Rotman has this to say: "First attempts to prove Corollary 7.23 were aimed at the polyhedron itself. For many years one tried to prove the Hauptvermutung (principle conjecture)...were this true there would be an easy provide of the topological variance of $H_*(K)$..." – Juan S Mar 9 '11 at 3:06

I didn't state it very well- what I meant is that standard algebraic topology textbooks take for granted (or cause the reader to take for granted) that topology of polyhedra is the same as topology of simplicial complexes. Sean Tilson's response is spot-on; I'll restate it in my own words.
By simplicial approximation, continuous maps of polyhedra are homotopic to PL maps. However, homeomorphisms of polyhedra might not be homotopic to PL homeomorphisms, merely to PL maps (the statement that they are is an equivalent formulation of the Hauptvermutung). So a (combinatorial) homotopy invariant of simplicial complexes might a-priori fail to be a homotopy invariant of polyhedra, unless one proves also that it's independent of the choice of simplicial approximation (which Matt E. and Qiaochu both say). It isn't enough just to show that it's independent under combinatorial homotopy of simplicial complexes.
I apologise for the confusion. My answer was essentially taken from Page 4 of The Hauptvermutung Book- I apologise also for the lack of attribution. I interpret what it says there as the assertion that textbooks don't tend to check independence with respect to choice of simplicial approximation; and my own experience (or inattentiveness) leads me to suspect that this is indeed the case. I'll look in a library on Monday, and maybe edit this response again.

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Do you mean that a PL-homeomorphism invariant might fail to be an unrestricted homeomorphism invariant? If not, what exactly is the distinction between a "homotopy invariant of simplicial complexes" and a "homotopy invariant of polyhedra"? – Matt E Mar 4 '11 at 5:39
@Daniel: An "anonymous user" is trying to edit your response, adding a paragraph at the end. If this is you, please log in, then you'll be able to edit it yourself without submitting it for approval. – Arturo Magidin Mar 4 '11 at 17:29
I merged your accounts. – Willie Wong Mar 4 '11 at 20:41
I see. Thank you for the clarification! – Qiaochu Yuan Mar 4 '11 at 22:31
Dear Daniel, Thanks for expanding on your answer. Regards, – Matt E Mar 5 '11 at 1:15

One doesn't need to explicitly compare with singular homology to get homotopy invariance (although that is certainly one way to do it), and one certainly doesn't need the Hauptvermutung (thankfully, since it is false in general). Rather, as you say, one can simplicially approximate a continuous map between simplicial complexes, and thus make simplicial homology a functor on the category of simplicial complexes and continuous maps (which, as you observe, will then factor through the homotopy category). Just thinking it over briefly, it seems to me that the hardest step in this approach will be to show that the induced map on homology really is independent of the choice of a simplicial approximation. Presumably this was covered in the lectures you are attending. (In fact, I guess this amounts to the fact that homotopic simplicial maps induce the same map on homology, which you say you have proved.)

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I deleted my answer because it was circular. To show simplicial homology is equal to singular homology by showing that it satisfies the Eilenberg-Steenrod axioms, you need to show it is homotopy invariant! Simplicial approximation should work without too much difficulty. – Grumpy Parsnip Mar 3 '11 at 11:31

I think maybe what he means is that people think that simplicial homology is a homotopy invariant of polyhedra, or at least take the fact for granted, not that simplicial homology is not a homotopy invariant of simplicial complexes. I guess his main point is that homotopy invariants of simplicial complexes are not necessarily homotopy invariants of polyhedra.

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I don't know what it means to say that something is a homotopy invariant of simplicial complexes if it doesn't mean that it's a homotopy invariant of their polyhedra. – Qiaochu Yuan Mar 3 '11 at 16:51
The point is that simplicial complexes, as topological spaces, are not the same as what Dan means by polyhedra. It appears that he means PL-manifolds, which I don't understand really except that I know they might not have a preferred triangulation, but they are triangulable. There is more structure flying around. However, every simplicial complex has an obvious preferred triangulation. – Sean Tilson Mar 4 '11 at 23:18