Showing that $\phi_1 + \phi_2 - \phi$ is a boundary We defined the standard $n$-simplex with ordering as $$ \sigma_n = \left\{ (t_0, \ldots, t_n) \in \mathbb{R}^{n+1} : \sum_i t_i = 1, t_i \geq 0 \right\}. $$ Then $S_n(X) $ is defined as the free abelian group whose basis is the set of all singular $n$-simplices of $X$. There is a boundary operator $\partial: S_n(X) \rightarrow S_{n-1} (X)$ given by $$ \partial = \sum_{i=0}^n (-1)^{i} \partial_i. $$ 
Problem: Consider a singular $1$-simplex $\phi: \sigma_1 \rightarrow X: (t_0, t_1) \mapsto \phi(t_0, t_1). $ Choose any $a \in \mathbb{R}$ with $0 < a < 1$ and define two other singular $1$-simplices as $$ \phi_1: \sigma_1 \rightarrow X: (t_0, t_1) \rightarrow \phi(t_0 + t_1 (1-a) , t_1 a) $$ and $$ \phi_2 : \sigma_1 \rightarrow X: (t_0 , t_1) \mapsto \phi(t_0 (1-a), t_0 a + t_1). $$ I have to prove that $\phi_1 + \phi_2 - \phi$ is a boundary. 
I'm not sure how to prove this. I already proved that a constant path is a boundary, and also that $\psi + \psi^{-1}$ (where $\psi^{-1}$ is the path traversed in opposite direction)) is a boundary. I think I may need to use this here, but not sure how. 
 A: I can give you the intuition behind the answer and someone else can fill in the exact details but here is the gist of it. Say you have some $1$-simplex $\phi$ in your space $X$:

You should picture $\phi_1$ and $\phi_2$ as the two $1$-simplices which join together to form $\phi$ like this:

Then if we can find a $2$-simplex $\sigma: \Delta^2 \rightarrow X$ such that $\partial_1 \sigma = \phi$, $\partial_0 \sigma = \phi_1$, $\partial_2 \sigma = \phi_2$ then it's easy to check that $\partial \sigma = \phi_1 + \phi_2 - \phi$. We construct $\sigma$ as the composition of these functions: 

The first map $f:\Delta^2 \rightarrow \Delta^1$ projects the top point of the triange to the point in $\Delta^1$ such that the red bit in $\Delta^1$ is of length $a$ thus "squishing" $\Delta^2$ leaving the bottom portion fixed and the second map is just $\phi : \Delta^1 \rightarrow X$.
You can intuitively see that $\sigma$ fits our criteria. Restricting $\sigma$ to the appropriate boundaries gives the maps we wanted and I have coloured in the appropriate colours to make this easier to see.
I'm pretty sure that $f$ is given by $(t_0, t_1, t_2) \mapsto (1-t_0 - (1-a)t_1,t_0 + (1-a)t_1)$ but can not verify it for myself so not 100% certain
