Let $\varphi \colon \sigma_1 \to X \colon (t_0,t_1) \mapsto \varphi(t_0,t_1)$ be a singular $1$-simplex, which is a path. The same path traversed in the opposite direction is then $\overline{\varphi} \colon \sigma_1 \to X \colon (t_0,t_1) \mapsto \varphi(t_1,t_0)$. I now have to show that $\varphi + \overline{\varphi} \in B_1(X)$.
However, in the previous exercise, I showed that for every $x_0 \in X$, $\varphi_{x_0} \colon \sigma_1 \to X \colon (t_0,t_1) \mapsto x_0$ is an element of $B_1(X)$. Because $\varphi$ is a path from $x_0$ to $y_0$, $\overline{\varphi}$ is a path from $y_0$ to $x_0$, so $\overline{\varphi} + \varphi$ is just a constant path and so we may conclude that the statement is true.
Is this reasoning correct or am I missing something rather important?
Best Answer
It's possible I screwed something up here, but I think this construction works for showing that a path plus its reverse is homologous to a constant, and is simpler than Max's more general answer to showing that in general $\alpha + \beta$ is homologous to $\alpha * \beta$ for any composable paths.
Represent points in $\sigma_2$ by their barycentric coordinates, i.e. they are points $(t_0, t_1, t_2)\in [0,1]^3$ such that $t_0 + t_1 + t_2 = 1$. Then define $\varphi_2\colon \sigma_2 \to X$ by
$$ \varphi_2(t_0, t_1, t_2) = \varphi_1(t_1, 1 - t_1) $$
where $\varphi_1$ is your singular $1$-simplex.
Then by definition of the boundary operator $\partial \varphi_2 = \partial_0\varphi_2 - \partial_1\varphi_2 + \partial_2\varphi_2$ where $\partial_i\varphi_2$ is $\varphi_2$ precomposed with the face map $l^i\colon \sigma_1 \to \sigma_2$, which are explicitly given by $l^0(s_0, s_1) = (0, s_0, s_1)$, $l^1(s_0, s_1) = (s_0, 0, s_1)$, and $l^2(s_0, s_1) = (s_0, s_1, 0)$. Therefore we can compute $\partial_0\varphi_2(s_0, s_1) = \varphi_1(s_0, 1 - s_0)$ which is just $\varphi_1$; $\partial_1\varphi_2(s_0, s_1) = \varphi_1(0, 1)$ which is constant and hence in $B_1$ by your exercise; and $\partial_2\varphi_2(s_0, s_1) = \varphi_1(s_1, 1-s_1) = \varphi_1(1-s_0, s_0)$ which is $\bar{\varphi}_1$. That is,
$$ \partial\varphi_2 = \varphi_1 - c + \bar{\varphi_1} \in B_1 $$