In Hatcher (p.$139$) the definition of cellular chain complex and its boundary $d_n$, is the following:
In my notes, I have an observation which states that $d_{n+1}$ in reality coincides with the boundary of the exact sequence of the triple $(X^{n+1},X^{n},X^{n-1})$. Is this true? I don't really see how could I prove it, since I think I can't use naturality of both boundary.
Any help or reference would be appreciated, thanks in advance.
On p. 118 Hatcher introduces the the long exact sequence of a triple $(X,A, B)$ $$\dots \to H_n(A, B) \to H_n(X, B) \to H_n(X,A) \stackrel{d_n}{\to} H_{n−1}(A, B) \to \dots$$ as the long exact sequence of homology groups associated to the short exact sequence of chain complexes formed by the short exact sequences $$0 \to C_n(A, B) \stackrel{i_n^{(X,A,B)}}{\to} C_n(X, B) \stackrel{j_n^{(X,A,B)}}{\to} C_n(X,A) \to 0 \tag{1}$$ The claim is that $d_n = J_{n-1} \circ \partial_n$, where $\partial_n : H_n(X,A) \to H_{n-1}(A)$ is the boundary of the long exact sequence of the pair $(X,A)$ and $J_{n-1} : H_{n-1}(A) \to H_{n-1}(A,B)$ is induced by quotient map $j_{n-1}^{(A,B)} : C_{n-1}(A) \to C_{n-1}(A) /C_{n-1}(B) $.
Given a short exact sequence of chain complexes $$0 \to U \stackrel{i}{\to} V \stackrel{j}{\to} W \to 0$$ the boundary $\partial : H_n(W) \to H_{n-1}(U)$ is defined as follows (see Hatcher p. 116):
Let $w ∈ W_n$ be a cycle. Since $j$ is onto, $w = j(v)$ for some $v ∈ V_n$. The element $∂v ∈ V_{n−1}$ is in $\operatorname{Ker} j$ since $j(∂v) = ∂j(v) = ∂w = 0$. So $∂v= i(u)$ for some $u ∈ U_{n−1}$ since $\operatorname{Ker} j = \operatorname{Im} i$. It turns out that $u$ is a cycle and that sending the homology class of $w$ to the homology class of $u$, i.e. defining $∂[w] = [u]$, produces a well-defined homomorphism.
Let us analyze this procedure for the sequence $(1)$ and the sequence $$0 \to C_n(A) \stackrel{i_n^{(X,A)}}{\to} C_n(X) \stackrel{j_n^{(X,A)}}{\to} C_n(X,A) \to 0 \tag{2}$$
Consider a cycle $w \in C_n(X,A)$ and get $v \in C_n(X)$ such that $j_n^{(X,A)}(v) = w$. Let $\bar v = j_n^{(X,B)}(v) \in C_n(X,B) = C_n(X)/C_n(B)$. Then $j_n^{(X,A,B)}(\bar v) = w$ since $C_n(B) \subset C_n(A)$. Find $u \in C_{n-1}(A)$ such that $i_{n-1}^{(X,A)}(u) = \partial(v)$. Let $\bar u = j_n^{(A,B)}(u) \in C_n(A,B) = C_n(A)/C_n(B)$. Then $i_{n-1}^{(X,A,B)}(\bar u) = \partial(\bar v)$ since $i_{n-1}^{(X,A)}$ induces $i_{n-1}^{(X,A,B)}$ on the quotients by $C_{n-1}(B)$. This shows that $$d_n([w]) =[\bar u] = [j_{n-1}^{(A,B)}(u)] = J_{n-1}([u]) = J_{n-1}(\partial_n([w]) .$$