Understanding boundaries in homology I'm trying to understanding how exactly I can construct a 2-simplex with the boundary $\bar\sigma + \sigma$ where $\sigma : \Delta^{1} = \{(t,1-t) \in \mathbb{R^2} \vert t \in [0,1] \} \rightarrow X$ and $\bar \sigma = \sigma (1-t)$ i.e. traversing $\sigma$ backwards. I realise that I probably want to find a way of sending $\Delta^2 \rightarrow \Delta^1$ and then act on $\Delta^1$ by $\sigma , \bar\sigma$, but I really am stuck and have no real way of visualising what's going on, any help would be hugely appreciated
 A: We can prove something more general: If $f$ and $g$ are homotopic paths in $X$, then $f-g$ is a boundary. From this, you can see how you might go about constructing your simplex.
The proof goes as follows, copying the idea from Lee's "Topological Manifolds". 
For convenience, take $\Delta_2$ to be the triangle with vertices $e_0=(0,0), e_1=(1,0)$ and $e_2=(0,1)$ and let $H:I\times I\to X$ be the homotopy from $f$ to $g$. Define $\alpha(x,y) = (x -xy,xy).$ Then, $\alpha$ maps the square onto $\Delta_2$ by sending each horizontal line $y=y_0$ in $I\times I$ to the segment $\overline {e_0(1-y_0,y_0)},$ that is, from the origin to a point on the hypotenuse of the triangle. 
it's easy to show that $\alpha$ is a quotient map: $\alpha $ is continuous and $I\times I$ is compact; and that $H$ is constant on the fibers of $\alpha:\ H(0,t)=f(0)$ and $\alpha$ is injective except on the left edge of the square, which it maps to $(0,0).$ So, we get a continuous map $\sigma:\Delta_2\to X$ given by $\sigma\circ \alpha=H.$
Now then, using the definition of boundary:
$\partial \sigma(s,t)=\partial \sigma(1-t,t)=\sigma(0,1-t,t)-\sigma(1-t,0,t)+\sigma(1-t,t,0).$
Changing to $x$ and $y$ and noting that they are just dummy variables, ranging from $0$ to $1$, the foregoing is
$\sigma(1-y,y)-\sigma(0,y)+\sigma(x,0)=\sigma(1-y,y)-\sigma(0,x)+\sigma(x,0).$ 
Putting this together with the homotopy $H$ and using the fact that  $\alpha(1,y)=(1-y,y)$, we have
$f(x)=H(x,0)=\sigma(x,0),\ g(x)=H(x,1)=\sigma(0,x),\ g(1)=H(1,y)=\sigma \circ \alpha(1,y)=\sigma(1-y,y).$
Thus, we have shown that $\partial \sigma=f-g+g(1).$ To finish, just observe that $g(1)$ is the boundary of the constant simplex $\tau:\Delta_2\to X:s\mapsto g(1).$ 
We have now proved that $f-g=\partial (\sigma-\tau)$ and so $f-g$ is a boundary.
The following picture has $f=f_0,\ f_1=g$ and $q=g(1)=f(1).$

