Is there any geometric motivation for the definition of the boundary operator $\partial:S_n(X)\rightarrow S_{n-1}(X)$?

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I started studying Singular Homology recently, and I stumbled upon this strange definition of the boundary operator in a singular n-chain, where $X$ is a topological space, and we have $\partial:S_n(X)\rightarrow S_{n-1}(X)$ given by

$$ \partial = \sum_{i=0}^n (-1)^i\partial_i $$

in which $\partial_i$ is the $i$th face operator, defined as

$$\partial_i\phi(t_0,\dots,t_{n-1}) = \phi(t_0,\dots,t_{i-1},0,t_i,\dots,t_{n-1}) $$

where $\phi $ is a singular $n$-simplex.

I tried to visualize what the boundary operator would result by applying it to the standard 1-simplex, 2-simplex and 3-simplex, but it hasn't given me any insights about its nature.

I understand that you can just take it as a definition, and use it to define the homology groups, and so on, but where did it come from? Does it have a geometric motivations, as the name suggests, or is it purely algebraic?

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Recall that the standard $n$-dimensional simplex $\Delta^n$ is a simplicial complex with one nondegenerate $n$-simplex $[id_{\Delta^n}]$. The $(n+1)$ face operators act on this simplex, sending it to its $(n+1)$ faces in dimensions $(n-1)$

Applying the formula for the boundary oeprator to $\Delta^n$, we see that on $[id_{\Delta^n}]$ it becomes the oriented sum of its $(n-1)$-dimensional faces.

Now if $\sigma$ is a $n$-simplex in $X$, then by naturality if should satisfy

$\partial\sigma=\sigma_*\partial [id_{\Delta^n}]$

so it sends $\sigma$ to its the oriented sum of its $(n-1)$-dimensional faces.