Simplicial Homology: The definition of cycles I'm trying to convince myself beyond a doubt that $n$-cycles should be defined as elements of $\ker \partial _n$. My intuition is along the lines of "a cycle is a boundary of some chain (not necessarily in the simplicial complex $K$), so all I really know is that its own boundary is zero because $\partial ^2=0$." My problem is that $\partial$ is itself only defined on chains of simplices in $K$, so really I'm not, in general, applying $\partial ^2$, and I can't justify the definition as I'd like.
What's a better way to carefully justify this definition?

Added:
I guess my phrasing was unclear, so I'll try to ask my question better. In Rotman's Introduction to Homological Algebra, the author says something along the lines of:

Some $n$-chains ought to be boundaries of a union of some $n$-simplices (not necessarily in the complex) - call these $n$-cycles. Some $n$-chains are boundaries of things in the complex - call these $n$-boundaries.

I'm trying to understand how the formal definitions fit this description. In the case of $n$-boundaries, it's clear why they should be defined as elements of $\text{im }\partial_{n+1}$, since that means - by definition - they are boundaries of some $(n+1)$-chain. What I don't understand, is how $\ker \partial _n$ captures the $n$-chains that are boundaries of "something" not necessarily in the complex (e.g boundaries of holes).
My question is how does $\ker \partial _n$ capture the right $n$-chains.
 A: Let's start with 1-cycles. 
The canonical example of a 1-cycle is an oriented simple closed curve $\gamma$ in the 1-skeleton of the simplicial complex, subdivided into 0-cells and oriented 1-simplices. Notice that each 0-simplex in the simplicial complex has exactly the same number of incoming oriented 1-simplices as it has outgoing oriented 1-simplices: that number is either 1 or 0 depending on whether the given 0-simplex is or is not a point on $\gamma$. It follows that $\gamma$ is in the kernel of $\partial$, in other words $\gamma$ satisfies the definition of a 1-cycle.
Now, one wants 1-cycles to be closed under addition and multiplication by coefficients. So, one wants to be able to add up a linear combination of arbitrary simple closed curves and still have a 1-cycle. Fortunately the kernel of $\partial$ is closed under arbitrary linear combinations so this works.
Finally, one can go one step further and show that anything in the kernel of $\partial$ can be rewritten as a sum of a finite number of oriented simple closed curves (using integer coefficients) or as a linear combination of oriented simple closed curves (using real coefficients).
To summarize, the definition of a 1-cycle is crafted so that the things that we think of intuitively as "1-cycles", namely the oriented simple closed curves, indeed satisfy the definition, and in fact they generate the whole group of 1-cycles in a very concrete sense. 
Having said that, one then steps away from the intuitive concept of "oriented simple closed curves" and adopts the definition of 1-cycles with glad heart, because really it is much easier to work with.

Now one can do something similar with 2-cycles. One might try to use embedded oriented 2-manifolds in the 2-skeleton as the canonical examples of 2-cycles. This does not quite work out, unfortunately; think of the 2-complex obtained by pinching the north and south poles of the 2-sphere to a single point. Instead, for the canonical examples can work with local embeddings of oriented 2-manifolds in the 2-skeleton. With a little work, a bit more than in the 1-dimensional case, one can show that each of these canonical examples is a 2-cycle, and that the general 2-cycle is a sum of 2-cycles each of which is one of the canonical examples.

This might be enough for you to now see how the definition of $n$-cycles is formulated, as an abstraction of the definitions when $n=1$ and $2$. One interesting and strange thing that happens, is that it is no longer true that one can use immersed oriented $n$-manifolds as the canonical examples of $n$-cycles, but that is a much deeper issue.
