Naturality of Bockstein Homomorphism with respect to Boundary Homomorphism Let $A \subset X$ be topological spaces and let $0 \to G' \overset{h}{\to} G \overset{k}{\to} G'' \to 0$ be a short exact sequence of abelian groups. We get a long exact sequence of homology groups:
$$\cdots \overset{\beta}{\to} H_n(X,A;G') \overset{h_\sharp}{\to} H_n(X,A;G) \overset{k_\sharp}{\to} H_n(X,A;G'') \overset{\beta}{\to} H_{n-1}(X,A;G') \to \cdots.$$
Here $h_\sharp$ and $k_\sharp$ are coefficient homomorphisms and $\beta$ is the Bockstein operator. I'm trying to show that $\beta$ is natural with respect to the boundary homomorphism of the pair $(X,A)$; that is, $\partial_*\beta = \beta\partial_* : H_n(X,A;G'') \to H_{n-2}(A;G')$. This should be a straightforward exercise in unpacking definitions, but I'm runnning into some trouble. Here's my attempt:
It suffices to verify that the maps commute on the generating set of elements of the form $[u \otimes x]$, where $u$ is a relative $n$-chain in $C_n(X,A)$ and $x \in G''$. Applying $\partial_*$, this element maps to $[\partial u \otimes x]$, where $\partial u \in C_{n-1}(A)$. Now we apply $\beta$: First we pick an element of $C_{n-1}(A;G)$ which maps to $\partial u \otimes x$ under $1 \otimes k$; the element $\partial u \otimes y$ will do the trick, where $y$ is any element such that $k(y) = x$. Now we apply the boundary map $\partial \otimes 1$ to this element, and here's where I run into trouble: doesn't this give $\partial^2u \otimes y = 0$? I tried performing $\beta$ and $\partial_*$ in the opposite order and did not get zero, so something's definitely wrong here. Where did I mess up?
 A: You seem to be using $\partial$ to mean both the connecting homomorphism and the boundary map of chain complexes.  While related, they are not exactly the same.  There are also several connecting homomorphisms in play here associated to the short exact sequences of complexes
\begin{equation}
C_\bullet(A; ?) \to C_\bullet(X; ?) \to C_\bullet(X,A; ?)
\end{equation}
\begin{equation}
C_\bullet(??; G') \to C_\bullet(??; G) \to C_\bullet(??; G''),
\end{equation}
where $? \in \{G', G, G''\}$ in the first line and $?? \in \{A, X, (X,A)\}$ in the second, and you should take care not to confuse them.
Anyway, here is a hands-on way of showing that the Bockstein commutes with the connecting homomorphism of pairs of spaces.  Assuming you've already shown that the chain level definitions are well-defined, you want to show that the following diagram "commutes":
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & & & & & \uparrow \\
C_n(X;G'') & & & & & & C_{n-1}(X;G') \\
\downarrow & & & & & & \downarrow \\
C_{n-1}(X;G'') & & & & & & C_{n-2}(X;G') \\
\uparrow & & & & & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}
For example, the top row should be read as "take a cycle in $C_n(X,A;G'')$, lift it to a chain in $C_n(X,A;G)$ using the short exact sequence of groups, apply the boundary map to get an element in $C_{n-1}(X,A;G)$, then use the short exact sequence of groups again to get an cycle in $C_{n-1}(X,A;G')$".  Upon passing to homology, this is the definition of the Bockstein homomorphism.  The other sides of the diagram should be interpreted similarly.
To show that the diagram "commutes", simply note that it is built out of smaller squares:
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_n(X;G'') & \leftarrow & C_n(X;G) & \rightarrow & C_{n-1}(X;G) & \leftarrow & C_{n-1}(X;G') \\
\downarrow & & \downarrow & & \downarrow & & \downarrow \\
C_{n-1}(X;G'') & \leftarrow & C_{n-1}(X;G) & \rightarrow & C_{n-2}(X;G) & \leftarrow & C_{n-2}(X;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}
You should check that each of the smaller squares commute.  Therefore, the large outer square "commutes" too, and this shows that the Bockstein and the connecting homomorphism commutes.

EDIT: Following the discussion in the comments, here is a simplified toy example that might make things clearer.  Consider the map of chain complexes:
\begin{equation}
\begin{array}{ccc}
0 & \rightarrow & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \xrightarrow{f} & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \rightarrow & 0
\end{array}
\end{equation}
where the maps $\mathbb{Z} \to \mathbb{Z}$ are all identities and everything else is $0$.  It's obvious that $\partial^2 = 0$ for both the chain complexes on the left and right, but if you start at the upper right corner, apply $\partial$, then $f^{-1}$, and then $\partial$, you would get the identity map, not zero.
