Let $f,g:\Delta_1\to X$ be two paths in $X$ such that $f(e_1)=g(e_0)$. Prove that $f*g-f-g\in S_1(x)$ is a 1-boundary. (Here $f*g$ denotes the concatenation of paths $f$ and $g$).
I have thought about solving this exercise in the following way:
Note that $$\partial(f*g-f-g)=\partial(f*g)-\partial(f)-\partial(g)=g(e_1)-f(e_0)-(f(e_1)-f(e_0))-(g(e_1)-g(e_0))=0$$ then $$f*g-f-g\in ker(\partial:S_1(X)\to S_0(X))$$, so $$[[f*g-f-g]]\in H_1(S_*(X))=\frac{ker(\partial:S_1(X)\to S_0(X))}{Im(\partial:S_2(X)\to S_1(X))}$$.
What I want to obtain is that $[[f*g-f-g]]=[[0]]$, because if the latter is true, then $f*g-f-g-0=f*g-f-g\in Im(\partial:S_2(X)\to S_1(X))=B_1(S_1(X))$ would have to be that $f*g-f-g\in B_1(S_1(X))$ and thus $f*g-f-g$ is a 1-boundary. How can I prove then that $[[f*g-f-g]]=[[0]]$? Is what I did fine? Thank you.