In the $1$-sphere $S^1$, let us pick a singular $1$-simplex $c$ with ends $x$ and $y$ on its equator, as the following image shows.
If we reverse the path $c$ and denote it by $c'$, then the $1$-chain $c+c'$ is a cycle, since $\partial_1(c+c')=x-y+y-x=0$.
If we define a singular $1$-simplex $c''$ that originates at $y$ and travels along $c$ to $x$ and travels back along $c'$ to $y$, then $c''$ itself is a $1$-cycle.
My question is:
Are $c+c'$ and $c''$ $1$-boundaries?
I think they are boundaries since the degrees of the maps $c+c'$ and $c''$ should be zero (intuitively, neither $c+c'$ nor $c''$ travels around the origin while the degree of a continuous $f\colon S^1\to S^1$ is the winding number of it), but I cannot actually figure out a proof that $c+c', c''\in B_1(S^1)$ (e.g., a proof that their degrees are both zero).
Thanks in advance...
Yes, they are boundaries. One way to prove this is that they are contained in the image $X$ of $c$, which is homeomorphic to an interval and hence contractible. So $H_1(X)=0$, and any $1$-cycle in $X$ is a boundary in $X$ (and thus also in $S^1$.