Empty boundary of a p-chain 
Definition: A $p$-cycle is a $p$-chain with empty boundary: suppose that $\phi$ is a group homomorphism from $C_{p}$ to $C_{p-1}$. Then,$\forall \alpha \in C_{p}, \phi(\alpha) = e_{+} \in C_{p-1}$.

Persistent Homology: An Introduction and a New Text Representation for Natural Language Processing
In the paper the author provides an example of an empty boundary 

I understand that a if boundary which maps elements in a $C_{p}$ group to the identity element in $C_{p-1}$, then the elements in $C_{p}$ are $p$-chains with empty boundary.
Can someone help me to understand why this is so? - the image on the left is an example of a 1 - cycle.
 A: I think in the notes you attached, the definition of a $p$-cycle is in Definition 18. Given a $p$-chain $\alpha$, its boundary is a $(p-1)$-chain $\beta$ such that $\partial\alpha=\beta$, where $\partial:C_p\to C_{p-1}$ is a boundary operator. Consequently, $\alpha$ has no boundary if $\partial\alpha=0$, where $0$ is the identity element of $C_{p-1}$ (we call it zero since in many contexts of homology theory, the operation of the group element is in additive notation).
The simplest way to make this intuition precise is to use "barycentric coordinates" of $R^d$, i.e. you label simplexes by its vertices as points in $R^d$. For our purposes let us use the diagram above.
For the left diagram, we have a 1-cycle (which is an "empty" triangle with no interior). Call this cycle $\alpha=(x_0x_1x_2)$ where $x_j$ label vertices. Note that its edges are also 1-chains, i.e. $(x_0x_1),(x_1x_2),(x_2x_0)\in C_1$. The boundary operator acting on an edge, for example $(x_1x_2)$, gives the two boundary "points", namely $x_2-x_1$. You could actually use $+$ instead of minus since following this paper, it is the sum modulo 2 ($+_2$), as in Definition 17. Following this, you can show that
$$\partial(x_0x_1x_2)=x_0-x_1+x_1-x_2+x_2-x_0 = 0\,.$$
This formalizes the intuition that "a closed loop has no boundary". Similarly, on the right diagram, the straight vertical right edge has two boundary points (in red) because of the definition above. This formalizes the idea that a finite line segment has two boundaries (namely, its endpoints).
