If $f$ is a (not necessarily closed) path in $X$, prove that the $1$-chain $f$ is homologous to $-f^{-1}$
The book is suggesting to use the fact that the Hurewicz map $\phi : \pi_1(X, x_0) \rightarrow H_1(X)$ is a homomorphism and suggests this graph as a hint.
I can see that $((f*f^{-1}) * f) * f^{-1} \simeq c \text{ rel $\{0,1\}$}$, where $c$ is a constant mapping. I know I have to show that cls $f \eta = $ cls $(-f^{-1}) \eta$, where $\eta : \text{1-simplex } \rightarrow I $. Or I think I have to show $f - f*f^{-1} + f^{-1} = 0$ with $f*f^{-1}=0$, but from here I'm stuck.
Anyone have any ideas?
The book you are using probably showed that when $f$ and $g$ are paths in $X$ such that $f(1) = g(0)$ and $f*g$ is the path "$f$ followed by $g$", then as simplices $$ f + g \sim f*g $$ which we express by saying that $f+g$ and $f*g$ are homologous and which means that $f+g-f*g$ is a boundary. This can be used to show that the Hurewicz map is a homomorphism. Can you see how to apply this to prove that $f \sim -f^{-1}$?