Two cycles are said to be homologous if their difference is a boundary (usual meanings implied).

What is the motivation behind this definition or the intuitive meaning it carries?

I am looking of something along the line of definition of homotopic curves, that they are curves which can be continuously deformed into one another.

Or may be if such a simple explanation isn't possible, what developments led to this particular definition, and what geometric properties of the topological space and homeomorphisms does it state.

Diagrams will be exceptionally helpful if possible.

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    $\begingroup$ This is an atypical reference, but pages 110-111 of this arXiv paper have a nice intuitive discussion. (The whole chapter is pretty good.) $\endgroup$ – Semiclassical Jul 28 '14 at 19:24
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    $\begingroup$ Some of the answers here mathoverflow.net/questions/640/… look helpful. $\endgroup$ – Conifold Jul 28 '14 at 19:26
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    $\begingroup$ This Wikipedia page on Pochhammer contours is also surprisingly relevant: it gives an explicit example of a contour which is homologous to zero but not homotopic to zero. (That means that you can't deform the contour into a point, but it still would integrate to zero because it has zero net winding around every point.) $\endgroup$ – Semiclassical Jul 28 '14 at 19:32
  • $\begingroup$ I'd start by getting a sense for what a cobordism between manifolds means. Then look at the definition of singular bordism and bordisms between singular maps. Next, think of a singular chain as a degenerate singular bordism. That's my preference. I think of homology between cycles as a triangulated version of bordism between cycles. $\endgroup$ – Ryan Budney Jul 28 '14 at 20:21

As mentioned by @RyanBudney, homology between cycles is like a PL version of cobordism between manifolds inside an ambient space. Cobordism just like homotopy, but continuous deformation of one submanifold to another instead of just loops.

(note : I have added hover texts with the pictures, please place your mouse point over the pictures to see the corresponding texts explaining them)

$M, N$ be closed $n$-manifolds. $W$ be a $(n+1)$-manifold such that $\partial W = M \sqcup N$. Then $W$ is said to be a (unoriented) $\it{cobordism}$ between $M$ and $N$, and $M$ is said to be $\it{cobordant}$ to $N$. One can easily verify that cobordism is an equivalence relation: reflexivity ($M \sqcup M$ bounds $M \times I$) and symmetry is obvious, and transitivity can be seen by gluing cobordisms along boundaries.

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Define $\Omega_n^{un}$ to be the collection of all cobordism classes of manifolds $[M]_{bor}$ with group operation $[M]_{bor} + [N]_{bor} = [M \sqcup N]_{bor}$. The identity under this group operation is the cobordism class $[\emptyset]_{bor}$ of the empty manifold. $\Omega_n^{un}$ is called the $n$-th unoriented cobordism group.

If $X$ is a simplicial/delta-complex, $\xi$ a simplicial $n$-cycle. Then $\xi$ can be thought, roughly, as a subcomplex of $X$ (by realizing each simplex in the formal linear combination as a simplex of the subcomplex, with products $n \cdot \Delta^i$ thought as an $i$-simplex thickened by a factor of $n$) with no boundary (i.e., no "open edges"). $\xi'$ be another $n$-cycle, realized as a subcomplex-ish thing.

If $\xi$ and $\xi'$ are homologous, the we know there is a $(n+1)$-chain $\sigma$ such that $\partial \sigma = \xi - \xi'$. But this $\sigma$ can also be realized as an $(n+1)$-dimensional subcomplex(-ish) of $X$ but with a few open edges this time such that boundary of the subcomplex is precisely $\xi - \xi'$. This reminds us of bordism, doesn't it?

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Indeed, more formally, if we work in the singular context, then an $n$-cycle $\xi$ can be realized as a map $\xi : K_\xi \to X$ as follows: realize $\xi$ as the formal linear combination $\sum_i \varepsilon_i e^i$ where $e^i$ are singular $n$-simplices and $\epsilon_i$ are $\pm 1$ (repeatation is allowed in the sum). Glue $n$-simplices appropriately according to how pair of singular $(n-1)$-simplices cancel in $\partial \xi = 0$ (there is a little subtlety here, but let's ignore that for now). We get the delta complex $K_\xi$ and the map $K_\xi \to X$ is obtained from sending each simplex to $X$ by the singular simplices $e^i : \Delta^i \to X$.

Thus, singular $n$-cycles are merely maps $\Delta \to X$ from delta complexes. Given two such maps $\xi : \Delta^n \to X$ an $\xi' : \Delta'^n \to X$, we see $\xi$ and $\xi'$ are homologous if $\xi$ and $\xi'$ extend to a map $\zeta : \mathbf{\Delta}^{n+1} \to X$ where $\partial \mathbf{\Delta}^{n+1} = \Delta^n \sqcup \Delta'^n$, i.e., $\zeta|_{\Delta^n} = \xi$ and $\zeta|_{\Delta'^n} = \xi'$. One can similarly check that this is an equivalence relation, and the equivalence classes of maps from certain delta complexes to $X$ forms precisely the $n$th singular homology group.

So homology between cycles in $X$ is merely (sort-of-)cobordism between simplicial complexes inside the ambient space $X$.

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