Proof and Intuition for $\partial_{q-1}\circ \partial_{q} = 0$ for $q\ge 2$ I am reading Singular Homology from Tammo Tom Dieck's Algebraic Topology. The author introduces the boundary operator $\partial_q$, and notes the boundary relation $\partial_{q-1}\circ \partial_{q} = 0$ for $q\ge 2$. I need help with the proof of this result and would appreciate it if anyone could provide some intuition behind it.
Here is the context (paraphrased) from the book (Pg. $224$, Section $9.1$, Chapter $9$: Singular Homology):

Let $\delta_i^n: [n-1]\to [n]$ be the (weakly increasing) injective map which misses the value $i$. This induces an affine map $d_i^n = \triangle(\delta_i^n): \triangle[n-1]\to \triangle [n]$, where $\triangle[n]$ denotes the standard $n$-simplex embedded in $\mathbb R^{n+1}$. Note that $\delta_j^{n+1}\delta_i^n = \delta_i^{n+1} \delta_{j-1}^n$ for $i < j$. This gives $d_j^{n+1}d_i^n = d_i^{n+1} d_{j-1}^n$ for $i < j$. A continuous map $\sigma:\triangle^n \to X$ is called a singular $n$-simplex in $X$. The $i$th face of $\sigma$ is $\sigma\circ d_i^n$. $S_n(X)$ is the free abelian group with basis the set of singular $n$-simplices in $X$. The boundary operator $\partial_q: S_q(X)\to S_{q-1}(X)$ is defined for $q\ge 1$ by $\partial_q: \sigma\mapsto \sum_{i=0}^q (-1)^{i} \sigma d_i^q$ and for $q\le 0$ as the zero-map. $\partial_{q-1} \circ \partial_q = 0$ is known as the boundary relation. For more details, see a screenshot from the book here.



*

*Proof:
Here is my attempt, which follows the hint provided in the book. We have \begin{align*}
\partial_{q-1}\circ \partial_{q}(\sigma) &= \partial_{q-1}\left(\sum_{j=0}^q (-1)^i \sigma d_i^q \right)\\ &= \sum_{j=0}^q (-1)^j \sum_{i=0}^{q-1} (-1)^i (\sigma d_j^q) d_i^{q-1}\\ &= \sum_{j=0}^q \sum_{i=0}^{q-1} (-1)^{i+j} \sigma d_j^q d_i^{q-1}\\ &=
\sum_{j=0}^q \sum_{i< j} (-1)^{i+j} \sigma d_j^q d_i^{q-1}  + \sum_{j=0}^q \sum_{i\ge j} (-1)^{i+j} \sigma d_j^q d_i^{q-1}\\
&=  \sum_{j=0}^q \sum_{i< j} (-1)^{i+j} \sigma d_i^q d_{j-1}^{q-1}  + \sum_{j=0}^q \sum_{i\ge j} (-1)^{i+j} \sigma d_j^q d_i^{q-1}\\ &\stackrel{?}{=} 0
\end{align*}
Why is this zero? I suppose some change of indices, etc. should show that the first sum is the negative of the second, but I'm not able to do it. Any help would be great.


*Intuition:  Why should $$\partial_{q-1}\circ \partial_{q} = 0$$ be true? I'm looking for intuition, and not formal proof (since that has already been discussed in the first half of this post.)

Let me know if any notation is not clear. Thanks a lot!
 A: Your proof is correct, every proof regarding boundary operators of homological theories is just a matter of signs, but how to understand the equality $\partial^2=0$, why is it called boundary operator?
As some comments suggested, you should work out for few simple cases to see what is happening here. For instance, let $X=[e_0,e_1,e_2]$ be a $2$-simplex (a triangle with vertices $e_0,e_1,e_2$), oriented counterclockwise.

Then $\partial(X) = [e_1,e_2] + [e_2,e_0] + [e_0,e_1]$ and you see that applying $\partial$ gives us three edges which is informally the boundary of $X$ with respect to the given orientation. Moreover,
$$ \partial^2(X) = e_1 - e_2 + e_2 - e_0 + e_0 - e_1 = 0.$$
If moreover you're interested de Rham cohomology then $\partial^2=0$ is a consequence of Schwartz's lemma, asserting that we can change the order of taking derivatives of a multivariable function provided the function is sufficiently nice. In both cases, the slogan here is
The boundary of a space (manifold, etc) is a space of the same kind without boundary.
In higher dimensional cases $[e_0,...,e_n]$, the first time you apply the boundary operator, $\partial: \sigma \mapsto \sum (-1)^k\sigma_{\mid [e_1,...,\hat{e_k},...,e_n]}$, you get all the faces opposite to vertices, afterwards, you obtain all "faces of faces" but remember each "faces of faces" appears exactly twice with opposite signs (like $e_0,e_1,e_2$ in our example) due to the orientability and hence all the things cancel out.
