Trying to visualize a polygon in a space $X$ From Rotman's Algebraic Topology:

A polygon in a space $X$ is a $1$-chain $\pi = \sum\limits_{i=0}^k \sigma_i$ where $\sigma_i(e_1) = \sigma_i(e_0)$ for all $i$.
By a theorem proven in the book, all polygons are $1$-cycles.

From the book: a $1$-cycle is essentially "a sum (union) of oriented $1$-simplices in $X$ that ought the constitute the boundary of some union of $2$-simplices in $X$."
But a polygon must have equal endpoints. So (as a simple example) is three disjoint circles in $\Bbb R^2$ considered a polygon? As in $\gamma = \sigma_1 + \sigma_2 + \sigma_3$ where each $\sigma_i \colon \Delta^1 \rightarrow \Bbb R^2$ is the $i$th circle.
Am I misunderstanding the idea or is there a simpler visual for this?
 A: You may have made a transcription error when copying the definition of polygon. The definition Rotman gives (at least by the 4th corrected printing, 1998) says 

A polygon in a space $X$ is a $1$-chain $\pi = \sum_{i=0}^k \sigma_i$, where $\sigma_i(e_1) = \sigma_{i+1}(e_0)$ for all $i$ (indices are read mod$(k + 1)$). 

In other words, the end of $\sigma_i$ is the start of $\sigma_{i+1}$. By "indices are read mod$(k+1)$" Rotman just means that $\sigma_k(e_1) = \sigma_0(e_0)$. Intuitively this should remind you of a cycle in a graph. 
Rotman's proof that a polygon is a cycle is terse, and he doesn't mention that the sum is telescoping. Specifically I mean
$$\begin{align}
\partial \pi &= \sum_{i=0}^k \partial\sigma_i\\
&= \sum_{i=0}^k(\sigma_i(e_1) - \sigma_i(e_0))\\
&=\sigma_0(e_1) - \sigma_0(e_0) + \sigma_1(e_1) - \sigma_1(e_0) + \dots + \sigma_k(e_1) - \sigma_k(e_0)\\
&=(\sigma_0(e_1) - \sigma_1(e_0)) + (\sigma_1(e_1) - \sigma_2(e_0)) + \dots + (\sigma_k(e_1) - \sigma_0(e_0))\\
&= 0
\end{align}$$
The definition of polygon implies that $\cup_i im(\sigma_i)$ is connected, and therefore three disjoint circles is not a polygon.
