I have been trying to understand what homology groups are "talking about," and now I am wondering if the following works as a definition of homology.

But first, some illustration of what it is inspired by: if $D\subset\Bbb C$ is a domain it seems elements of $H_1(D)$ can be interpreted as contours to integrate over, considered with orientation (sign) and multiplicity. If $\oint_\gamma f(z)dz=0$ for all of the functions $f\in{\cal O}(D)$ (i.e. holomorphic $f:D\to\Bbb C$) then we treat $\gamma$ as $0$ in $H_1(D)$. Thus, loops are trivial if they can be contracted to a point within the domain $D$. On the other hand, if a loop goes around a punctured point in $D$, there will be holomorphic functions in ${\cal O}(D)$ with a simple pole at the point and integrating such a function over the contour will yield a nonzero value.

There are some more relations satisfied by these contours-with-multiplicity. If $-\gamma$ is the same contour as $\gamma$ but traversed in the opposite direction then $\oint_{-\gamma} f(z)dz=-\oint_\gamma f(z)dz$ and hence $\oint_{\gamma+(-\gamma)}f(z)dz=0$ for all $f$, so we may conclude $\gamma+(-\gamma)=0$. And if $\gamma_1$ and $\gamma_2$ go around two distinct punctures respectively (and no others) with the same orientation then $\gamma_1+\gamma_2$ is equivalent to any contour that loops around *both* punctures (but no others) in the same direction. Without loss of generality I think we can consider $H_1(D)$ to be generated by simple loops.

To generalize to $n$ dimensions, we need vector calculus and differential forms to replace complex analysis. From what I understand, Stokes' and de Rham's theorems say that elements of homology are oriented domains of integration for differential forms counted with multiplicity. (I have not delved more deeply than this intuitive understanding, though.) In that spirit, I want to define my chain groups $C_n({\cal M})$ as $\Bbb Z$-linear combinations of closed, compact, oriented $n$-dimensional submanifolds, subject to the following two relations:

- If $\cal A, B$ intersect only on their boundary and $\cal A\cup B$ has a consistent orientation then we identify $\cal A\cup B$ with $A+B$. I believe an equivalent condition is that if $\cal A\cap B\ne\varnothing$ and have the same restricted orientation on $\cal A\cap B$ then $\cal A+B=(A\cup B)+(A\cap B)$.
- If $\cal C,C'$ are the same submanifold but with opposite orientations, then $\cal C'=-C$.

One may take the usual topological boundary (in which orientation is inherited) and extend linearly to obtain the boundary maps $\partial_n:C_n({\cal M})\to C_{n-1}({\cal M})$. Then $\partial^2=0$ and we may define the homology group $H_n({\cal M})$ as generated by the boundaryless chains modulo boundaries in the usual fashion ($H_n=\ker(\partial_n)/{\rm img}(\partial_{n+1})$). **Does this work as a viable definition?**

Just as with contours in complex analysis, it seems we can homotope any submanifold and still represent the same element of homology. (For instance, consider a sphere $\cal A$ around the origin in $\Bbb R^3-0$ with orientation chosen to make outward-pointing normals by the right-hand-rule. Say we homotope this to a smaller sphere $\cal B$ within $\Bbb R^3-0$. Then $\cal A-B$ will be the oriented boundary of some oriented annular region and hence $\cal A-B\equiv{\rm 0}~\Rightarrow A=B$ in $H_2(\Bbb R^3-0)$.

If we put a nice CW complex structure on $\cal M$, it feels like any submanifold (a generator) in the chain group $C_n({\cal M})$ may be homotoped to one comprised entirely of oriented cells from the complex. Thus, using CW complexes should represent a "discretization" of the definition I'm submitting that allows for finite computations to be humanly carried out. Moreever, every compact manifold can be triangulated, so it seems we can also relate it to singular homology.

So, anyway, to repeat my question: does this work as a viable definition of homology? Does this idea have a name? Does it seem more natural than other definitions? Are there technical obstacles to relating it to usual definitions of homology? If this doesn't work, why not?

One idea offered in chat is that I may not be able to homotope (or isotope, whatever may be necessary to make things work out) two knots in a way that can be broken apart into annuli (if annuli bound the before/after submanifolds, that would make them all equal in $H_1$). The moral of that lesson is that we might need to introduce a third relation in $C_n({\cal M})$, that isotopic submanifolds are equal.