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.
Let $X$ be simply connected and let $ \tau $ be a cycle in $X$. If $ \tau(0)=\tau(1)=x_0$ and $\gamma $ is a path between $x\in X$ and $x_0$, is $\tau$ homologous to $ \gamma\tau\gamma^{-1} $?  I think it is, but I'm not sure about the proof. Can I use the fact that $ \gamma\tau\gamma^{-1} $ is homologous to $ \gamma+\tau+\gamma^{-1} $, where $+$ is the operation between 1-chains (that is abelian)?
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In particular, there have been a few posts where someone has made a substantial edit to a question or answer which significantly alters its meaning. Typically I'd reject that as "too radical", since it can't be known that the new meaning of the post was what the original author intended.  However, in some cases it is clear from the comments that the original author is aware of the intended change, and as such the editor is not in fact misrepresenting the author's intentions. In these cases, it would have been useful to see the comments when making the review decision.  As an example, a common cause of this is when an inexperienced user is asked to clarify their question, and ends up writing a comment that would be better off as an edit to the question (due to length, excess code content, etc), presumably due to lack of awareness that the question can be edited. 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