# Can we make rigorous the 'obvious' characterisation of singular homology?

It is a well known and often touted fact that the singular homology groups 'count the k- dimensional holes' in a space (see: How does singular homology H_n capture the number of n-dimensional "holes" in a space? for an explanation), but has anyone succeeded in making this heuristic rigorous?

I am aware of the work of Rene Thom on a related subject (see: Cohomology and fundamental classes), but am dubious that this comes close to what I am asking.

Specifically what I am asking is whether there is or indeed ever could be a theorem of the form:

Suppose $M^n \subset \mathbb{R}^m$ is a hausdorff [insert category here] manifold of dimension n: let $Y_k$ be the set of generators of $H_k(M^n)\otimes \mathbb{Q}$ with $k\neq 0,n$ then for each $[x]\in Y_k$ $\exists x \in [x]$ and a point of $x$, $\hat{x}$, s.t. $\exists$ a convex neighbourhood $U$ of $\hat{x}$ with $U \cap M$ homotopic to $\mathbb{R}^{k+1} \setminus 0$.

(The torsion has been killed in an effort to make this a reasonable request)

If not, why not? (I am aware that there could be an arbitrary number of twists in the codimension of the fundamental class for example, but surely this could be sorted out!) Are there counter examples? Is there a situation where this is true?

Edit: After a counter example to the original conjecture, rather than retreating all the way back to manifolds Let's attempt to defend the intermediate bridge of $U \cap M$ homotopic to $S^{i_1} \times... \times S^{i_p}$ with $\Sigma_j i_j=k$

Edit2: $H_2(\mathbb{R}^3 \setminus T \wedge T)$ scuppers that. Compact orientable smooth manifolds it is then...

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My question that you cite above was not taken really seriously when I asked it. At least that was my impression. In any case, Mariano's answer to that question and the comments of Sammy Black and others to that question could be helpful. – Akela May 25 '10 at 11:55
The comments you refer to are certainly interesting- but they all seem to be either moral arguments for the above or constructions of rigorous frameworks for those moral arguments. What I am after is the rigorous argument itself. – Tom Boardman May 25 '10 at 13:03
What do you mean by twists in the codimension? the codimension is an integer. – Thomas Kragh May 25 '10 at 13:58
Tom, the condition you have formulated doesn't depend on the homology class $[x]$ (assuming $M$ connected), because any cycle is homologous to one whose image contains a given point. It's just a condition on the embedding and $k$. And the intersection of a convex $U$ in $\mathbb{R}^3$ with a standard embedded $S^2$ can be an annulus, even though $H_1(S^2)=0$. So true or false, I don't think this condition captures homology. – Tim Perutz May 25 '10 at 14:02
This question is unclear. Moreover, it's not clear to me why `obvious' is in the title. – Ryan Budney Jun 2 '10 at 14:31

Here is a counterexample: A torus embedded in $\mathbb{R}^3$ has a fundamental class in $H_2$, but there are no points with neighborhoods homotopy equivalent to $S^2$.

It might be interesting to try to formulate a condition for the homology of pairs $(X,Y)$.

Edit: I think you may want to change your conjecture by demanding that there exist neighborhoods of a certain type that represent a set of homology cycles that span $H_k$. Otherwise, there doesn't seem to be a relationship between the neighborhoods and the elements of homology. It is also not clear why convexity makes an appearance - you may want to demand the neighborhood be contractible in $\mathbb{R}^m$, though.

With that in mind, here is a family of counterexamples that actually satisfy the conditions you specified: Embed the torus $(S^1)^n$ into some $\mathbb{R}^m$, for $n>2$ and sufficiently large $m$. Then for $k$ ranging from $2$ to $n-1$, the rational homology in degree $k$ has dimension $\binom{n}{k}$, but no cycles represented by spheres (or punctured Euclidean spaces).

In general, I think you should change "punctured Euclidean space" to "compact connected orientable manifold". Such objects look like holes if they aren't boundaries of higher-dimensional manifolds (and if you squint enough).

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Yeah, saw that one coming- that was why I put $k \neq n,o$ in my dummy theorem, I'm fairly sure most H_n classes don't satisfy my condition, for that matter it doesn't make sense for H_o- it's the stuff in the middle that I think this might apply to. – Tom Boardman May 25 '10 at 15:00
Yeah, the neighbourhood thing is a real doozy, am trying to fix that (see Tim's comment/my reply above). As for the counter-example- Yes!!!! That does break it... maybe let's go for homotopy equivalent to a product $S^{i_1}\times... \times S^{i_p}$ $\Sigma i_j=k$ – Tom Boardman May 25 '10 at 16:10
Have your upvote back, and another for your troubles!! Oh, P.S. covexity was there to stop us just thickening an arbitrary k sphere and calling it a neighbourhood (or else the conjecture would hold trivially)- suppose contractible would be easier though – Tom Boardman May 25 '10 at 17:22

I think that your theorem is "almost" true if we restrict ourselves to a neighborhood of x in (some quotient of) the k-skeleton. And indeed, this is a way to give a precise meaning to "holes" in singular homology. Let me be more specific.

Let's consider a finite CW-complex. The space is built starting with a finite number of points, and attaching a finite number of cells of various dimensions. The k-skeleton $X_k$ is the union of all cells of dimension less than or equal to k.

The smash $X_k|X_{k-1}$ is obtained from $X_k$ by identifying all points in $X_{k-1}$. If X is a CW-complex, that "smash" is homeomorphic to a bouquet of k-spheres. These spheres are the "holes" we are looking after. From the standard identification $H_k(X_k|X_{k-1})\simeq H_k(X_k,X_{k-1})$ , we see that each sphere gives rise to an element in the relative homology group $H_k(X_k,X_{k-1})$. There are two ways these elements may fail to give an element in $H_k(X)$.

--> Instead of capturing a k-dimensional "hole", the cell may in fact "fill" a (k-1)-dimensional "hole". That's what happens when, for example, we cap a cylinder with a disk. So, we are only interested in elements in the kernel of the boundary operator $\delta_k : H_{k}(X_k,X_{k-1})\rightarrow H_{k-1}(X_{k-1})$.

--> The k-dimensional "hole" may be filled by some $k+1$-dimensional cell. So we should quotient $H(X_k,X_{k-1})$ by the image of the operator $H_{k+1}(X_{k+1},X_k)\rightarrow H_{k}(X_k)\rightarrow H_{k}(X_k,X_{k-1})$. Let us denote that image by $E_k$.

And it works. The group $H_k(X)$ is actually isomorphic to the quotient $ker\ \delta_k / E_k$. So there is a subset of the "holes" in $X_k|X_{k-1}$ that provide a generating family for $H_k(X_k)$. Arguably, the spherical neighborhood you are looking for exists in the smash, not in X, but still, I think it succeeds in making our intuition rigorous. As a reference, I may point to Greenberg "Algebraic topology, an introductory course" (21.8 ff).

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