Usually we define singular homology via the complex of singular chains: $$C_\ast(X)=\bigoplus_{n\geq 0}\mathbb Z\langle\sigma:\Delta^n\to X\rangle$$ where the right hand side denotes the free abelian group on the set of maps $\sigma:\Delta^n\to X$.

Suppose that we want the flexibility to use more general manifolds with corners in place of standard simplices. So, we define an alternative chain complex: $$C_\ast'(X)=\bigoplus_{n\geq 0}\mathbb Z\langle f:M^n\to X\rangle/\sim$$ where the right hand side denotes the free abelian group on the set of isomorphism classes of maps $f:M^n\to X$ where $M^n$ is a smooth oriented manifold with corners, modulo the relation that reversing orientation negates the generator (and also restricting to maps $f:M^n\to X$ for which reversing the orientation on $M^n$ gives a non-isomorphic map, just so we don't have a bunch of $\mathbb Z/2$'s floating around). The boundary map is clear: take the sum of all the "faces" of $M^n$ (this is ok, even though some of the faces of $M^n$ may not be embedded in $M^n$).

There is an obvious map $C_\ast(X)\to C_\ast'(X)$. Is it a quasi-isomorphism? My guess would be that it is, but I also found I was unable to prove it after thinking for quite a while. Surprising as it may sound, I don't even know how to, given a cycle in $C_\ast'(X)$, construct the homology class in $H_\ast(X)$ it is "supposed" to represent!

There's an obvious strategy of proof, namely "triangulate the manifolds with corners". However doing this compatibly with the boundary map, even for a specific cycle in $C_\ast'(X)$, seems hard.

I've tagged this "reference request" since it seems that a positive answer to my question is a "folk theorem". For instance, some quick searching revealed a paper http://arxiv.org/abs/math/0509532 which claims in Theorem 5 that (a slightly different version of) $C_\ast'(X)$ calculates singular homology. However all they say for the proof is that manifolds with corners can be triangulated (which, on its own, does not seem to me to be sufficient).

Another related construction in the literature is that of Max Lipyanskiy (http://www.math.columbia.edu/~mlipyan/SingHom.pdf), who shows that if we are only interested in the case $X$ is a smooth manifold, then (something very close to) the analogous complex built out of isomorphism classes of smooth maps from smooth manifolds with corners to $X$ does indeed calculate singular homology. Unfortunately, his proof of this fact uses smoothness in an essential way, so I do not see how to extend it to the case I am interested in.

  • $\begingroup$ This seems perfectly sufficient: just don't try to construct a chain homotopy but rather handle cycles one by one. And then it seems obviously injective and obviously surjective. $\endgroup$ – Alex Degtyarev Feb 16 '14 at 22:10
  • 2
    $\begingroup$ It would probably be easiest to just verify the Eilenberg-Steenrod axioms for "manifolds with corners homology" (I think you can just follow the usual proof for singular homology nearly word-for-word). This will show you that your homology theory is isomorphic to ordinary singular homology. The fact that the map you wrote down is the isomorphism should just follow by meditating on naturality and the proof that homology can be calculated using cellular homology. $\endgroup$ – Andy Putman Feb 16 '14 at 22:26
  • 1
    $\begingroup$ I mostly second Andy's suggestion: if it's a quasi-isomorphism locally then it should be globally by abstract nonsense, e.g. sheaf theory. However, I'm not used to thinking so much about manifolds with corners. If $X$ is a point, and $f:M\to X$ is the unique map and represents a cycle, what is $M$ the boundary of? If we were doing bordism, the answer might be "nothing", whereas simplices can always be coned off and you still have simplices. What's the analogous situation for manifolds with corners? $\endgroup$ – Greg Friedman Feb 17 '14 at 4:27
  • 2
    $\begingroup$ I think I spoke too fast when I wrote that one can following the usual proof word-for-word; I had quickly gone through it and didn't see any obstructions, but you're right that it's not so clear. But if it's true, then surely that's the way to go about proving it. $\endgroup$ – Andy Putman Feb 17 '14 at 5:05
  • 2
    $\begingroup$ @John: I'm a bit leery of your statement "I'm sure it's not so hard to show that if $M^n$ ($n>0$) is a closed manifold, then it is null-homologous in $C′_\ast(pt)$." This won't be true using just manifolds (as the bordism groups of a point are non-trivial), and I don't see immediately why allowing corners helps. For example, what would be the "null-homology" of the map from $CP^2$ to a point? $\endgroup$ – Greg Friedman Feb 18 '14 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.