Singular Homology: every $0$-chain is a $0$-circle

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I am having trouble understanding this fact, that is deemed as trivial and thus not proved in most books.

First of all, I understand that the boundary operator $\operatorname{Bdy}$ goes from $S_n$ to $S_{n-1}$ (where $S_n$ is the group of $n$-chains). In this case, which is the codomain? I assumed it is $S_0$ as well, exceptionally.

Secondly, if $f$ is a $0$-chain, it should have the $0$-simplex, a point, as it's domain. $\operatorname{Bdy}(f)$ should be $f(0)$ in this case. But $0$ does $0$ belong to the $0$-simplex?

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The codomain is $0$ literally, this also makes every $0$-chain into a $0$-cycle(respectively, if you like to think about it that way: a point has no boundary, if you think about it, every point is a degenerate simplex with trivial boundary).

I know this is quite a technicality, however, it helps a lot although sometimes it is also useful to pass to the reduced homology, where you kill the class of the basepoint, since precisely this technicality can make stuff weird.