Cycles and Closed Paths $\newcommand{\Im}{\operatorname{Im}}$

Show that if $\Delta$ is a triangulation, then $[a_1, b_1]+[a_2, b_2]+\cdots+[a_n,b_n]$ is a $1$-cycle precisely when the indicated oriented edges $1$-faces form a closed path.
This is what I have in my notes:
Out of a triangulation, we get a sequence $$0 \stackrel{\partial_0}\longleftarrow C_0 \stackrel{\partial_1}\longleftarrow C_1 \stackrel{\partial_2}\longleftarrow \cdots \longleftarrow C_n \stackrel{\partial_n}\longleftarrow \cdots$$
with $Z_n = \ker \partial_n \supseteq B_n = \Im (\partial_{n+1})$

If the oriented $1$-faces form a closed path, then they must be a boundary of the $2$-faces, but the boundary operator applied to $B_1$ is zero, since $\ker \partial_n \supseteq B_n = \Im (\partial_{n+1})$. So $\Im(\partial_2) = B_1  \subseteq \ker \partial_1$.
For the converse, we have: $1$-cycle $\implies$ boundary is zero $\implies$ is contained in the kernel of $\partial_1$. So now I have to prove that it is also contained in the image of $\partial_2$. Now I'm a bit confused, because the image of $[a,b,c]$ is $[a,b]-[a,c]+[b,c]$ ... so if a $1$-cycle is of that form, then it definitely is an image of a $2$-face. But we can have all kinds of $1$-cycles, right? For example $[a,b]+[b,a]$ is also a $1$-cycle and I can't really see how that would be a boundary of a $2$-face, since it is not of the form $[a,b]-[a,c]+[a,c]$.
Also, is there a formal definition of a "closed path"? I know what it means intuitively, because our professor showed it using a diagram, but I'm not sure if there is a formal definition, of if it's meant to only have an intuitive definition. I tried to google it but couldn't find anything useful.
 A: I think you may be mixing up the definitions of the various terms (chains, cycles, boundaries).
If $u$ is an $n$-chain and $\partial u=0$ we call $u$ a cycle. If there exists an $(n+1)$-chain $u'$ such that $\partial u'=u$ then we call $u$ a boundary. In homological algebra, we're particularly interested in the $n$-chains which are cycles but not boundaries, because such chains generate the $n$th homology group.
Let's go through an example. Let $\mathcal{K}$ be a simplicial complex with


*

*$0$-simplices $\{a,b,c,d\}$,

*$1$-simplices $\{[a,b],[b,c],[c,a],[a,d],[d,c]\}$, and

*$2$-simplicex $\{F=[a,b,c]\}$.


Draw a picture of this simplicial complex.
Now, let's find a couple of $1$-cycles in $\mathcal{K}$. It's easy enough to see that $\partial_1 ([a,b]+[b,c]+[c,a])=\partial_1(u_1)=0$ and also $\partial([c,a]+[a,d]+[d,c])=\partial(u_2)=0$ and so both $u_1$ and $u_2$ are cycles.
Now, we can also see that $\partial_2 (F)=u_1$ and so $u_1$ is also a boundary, as well as a cycle. Seeing as $C_2$ is generated by a single element (namely $F$), the image of a $2$-chain under $\partial_2$ will always be a linear combination of $u_1$. It follows that $u_2$ is not in the image of $\partial_2$ as $u_2$ can not be written as a multiple of $u_1$ and so $u_2$ is not a boundary, even though it is a cycle. This is pretty clear from the image you draw because it does not bound any $2$-simplex.
Given this example, I think you might be a bit better equipped to answer the question.
