This is a really simple question that I think I have answered, but I'm not altogether satisfied and would like confirmation or an alternative. We define the simplices
$$ \begin{align} [A,B] : \Delta^1\to\mathbb R & & [B,A]:\Delta^1\to\mathbb R \\ e_1 \mapsto A & & e_1\mapsto B \\ e_2\mapsto B & & e_2 \mapsto A \end{align} $$
where $\Delta^1$ is the standard 1-simplex and $e_1 = (1,0), \, e_2 = (0,1)$. I'm now asked to prove that $[A,B] \neq -[B,A]$ in $S_1(\mathbb R)$, but they are in fact equal in homology. Oddly the second part is the easiest, since I have found a chain $\sigma$ such that $\partial\sigma = [A,B] + [B,A]$. I'm still getting used to the group of simplicial chains just being declared as it is (unlike homotopy groups which are more obviously tied to topological operations), which is my confusion with the first part. Am I correct in saying that $[A,B] \neq -[B,A]$ because, since $S_1(\mathbb R)$ is the free abelian group generated by all 1-simplices, then $\langle [A,B]\rangle \cap \langle [B,A]\rangle = 0$ (since they are not equal as maps) and therefore one cannot be a multiple of the other?