TITLE: On defining homology groups
QUESTION [12 upvotes]: 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.

REPLY [6 votes]: Let me try to take your proposal seriously and see what happens. Your setting is the one of a smooth (everything will be smooth from now on) manifold $M$, where the group of  $k$-cycles is the free abelian group with the basis given by oriented $k$-dimensional closed submanifolds $c\subset M$ and the subgroup of boundaries is (essentially) generated by boundaries of compact $k+1$-dimensional submanifolds in $M$. 
Note. You are taking a further quotient given by identification $-c=\bar{c}$, where $\bar{c}$ is an oriented manifold equipped with the opposite orientation. I will keep this in mind. 
The result is an abelian group which I denote $h_k(M)$, but the question what is it good for and can you compute it? 
Everything is clear if $dim(M)=1$. After a bit of work (you will need to prove, among other things, smooth Jordan curve theorem in $R^2$), you prove that for oriented surfaces $M$, $h_k(M)\cong H_k(M)$, where $H_k$ is the singular homology group.  Then you try $M=RP^2$ and you realize (to your great distress) that $h_k(M)$ has infinite rank of cardinality continuum! Independent elements of $h_1(M)$ are represented by projective lines on $RP^2$. That's not good, since maybe you are hoping for an analogue of Euler's formula (Euler characteristic equals to the alternating sum of dimensions of homology group). You might declare at this point that you are interested only in oriented manifolds. Fine. Let's move to dimension 3. You prove (with some difficulty) that $h_k(S^3)$ is isomorphic to $H_k(S^3)$, so nothing surprising here. However, in dimension 4 you look at $S^4$ and realize that you have no idea what to do. Does $h_2(S^4)$ have finite rank or not? (I do not know either but I did not spend much time thinking about it.) 
Edit. Mike Miller noted in a comment that every surface in $S^4$ will bound a submanifold, so $h_2(S^4)=0$. Moreover, whenever $H_{n-2}(M)=0$, $h_{n-2}(M)=0$ as well, for every oriented $n$-dimensional manifold $M$.
Then you look at $CP^2$ and, again discover that $rank(h_2(CP^2))$ has cardinality of continuum! (Now because of complex projective lines.) At this point you might declare that you are only interested in homology of domains in $R^n$. Fine: You then find that $R^8$ contains a  copy of $CP^2$ which is nontrivial as an element of $h_4(R^8)$. (Since $CP^2$ does not bound any compact 5-manifold.) 
At this point you probably are ready to give up on using embedded manifolds as your cycles and boundaries. After all, in order to integrate differential forms, embeddedness is irrelevant. You use smooth maps of oriented manifolds (with boundary) into $M$ in order to define cycles and boundaries. This leads you to rediscover the bordism groups $O_k(M)$ of $M$. These are  interesting groups which are discussed in other answers. However, even the manifold consisting of a single point has nontrivial bordism groups (already $O_4(\{p\})\ne 0$), again because of $CP^2$. You may not like it either and decide to allow "smooth manifolds with mild singularities" for your chains. After all, you do not need smoothness in order to integrate. What constitutes "mild singularities" is debatable. The classical singular homology theory provides one answer. Another answer appears in the work of Mattias Kreck "Differential Algebraic Topology", which might be what you are searching for.  In the end, it yields groups isomorphic to the singular homology groups, see this paper by Kreck and Singhof. Sullivan's "Infinitesimal Computations in Topology" is also relevant here. 
Lastly, as for the request for 

"an answer that addresses the technical problems in this proposal and makes it either a equivalent to a well-known existing approach, or else superior to other approaches", 

it is akin to a request to show that a turtle either has equal speed or is faster than a hare.