Topology and Geometry, $1$-simplex is $2$-boundary consider page 172-173 in Topology and Geometry by Glen Bredon. In particular, Lemma 3.1.
The result is: If $f,g$ are paths in $X$ such that $f(1)=g(0)$ then the $1$-chain $f\cdot g - f - g$ is a boundary. Here, $f\cdot g$ is the concatentation of the paths $f$ and $g$.
He proves the result as follows:
On the standard $2$-complex $\Delta^2$, put $f$ on the edge $(e_0,e_1)$ and $g$ on $(e_1,e_2)$. Then define a singular $2$-simplex $\sigma: \Delta^2\rightarrow X$ to be constant on lines perpendicular to $(e_0,e_2)$ and $f\cdot g$ on the edge $(e_0,e_2)$.
Then the constructed $2$-simplex $\sigma: \Delta^2 \rightarrow X$  is continuous.
I also don't completely understand the construction. May someone elaborate? On the lines perpendicular to $(e_0,e_2)$, what is $\sigma(t_0,t_1,t_2)$? Is it continuous by the gluing lemma?
May someone elaborate on why $\Delta^2$ is continuous please?
 A: The relevant picture should be something like this:

This displays a 2-simplex and labels each point by where it goes under the map $\sigma$. On the short vertical line toward the left, the value will be constant at $f(1/4)$. On the center vertical line, the value will be constant at $f(1)=g(0)$. Each vertical line (each line perpendicular to the base of the triangle) will get sent to a constant, to either $f(t)$ or $g(t)$ for the appropriate value of $t$.
Edit: for continuity, one approach is to follow @LeeMosher's comment. Let's parametrize this 2-simplex not as a the standard 2-simplex but as follows: put the lower left corner at $(-1,0)$, the lower right corner at $(1,0)$, the top corner at $(0,1)$. Then define a map $\sigma$ from this triangle to $X$ by
$$
\sigma(s,t) = \begin{cases}
f(1+s) & \text{if } -1 \leq s \leq 0 \\
g(s) & \text{if } 0 \leq s \leq 1.
\end{cases}
$$
Since $f(1) = g(0)$, this is continuous. Compose with a linear homeomorphism $h$ from the standard 2-simplex $\Delta^2$ to this one to get a map $\sigma \circ h: \Delta^2 \to X$.
