The Bockstein is independent of lift and representative

The cochain $a$ in [F1] is a cocycle. [given, F1] Indeed, $i_{\ast }(\delta a)=\delta (i_{\ast }a)={\delta }^{2}b=0$; injectivity of $i_{\ast }$ gives $\delta a=0$.

Changing only the lift changes $a$ by a coboundary. [F1, step 1.1] If $b^{\mathrm{\prime }}$ is another lift of $c$, then $q_{\ast }(b^{\mathrm{\prime }}-b)=0$. Exactness gives a unique $h\in C^n(X;A)$ with $b^{\mathrm{\prime }}-b=i_{\ast }h$. If $a^{\mathrm{\prime }}$ is defined from $b^{\mathrm{\prime }}$, then $i_{\ast }(a^{\mathrm{\prime }}-a)=\delta (b^{\mathrm{\prime }}-b)=i_{\ast }(\delta h)$, hence $a^{\mathrm{\prime }}-a=\delta h$.

Changing the cocycle representative also changes $a$ by a coboundary. [F1, F2, step 2.1] Write $c^{\mathrm{\prime }}=c+\delta u$. Use the lifting choice of [F2] to take $v\in C^{n-1}(X;B)$ with $q_{\ast }v=u$. For an arbitrary lift $b^{\mathrm{\prime }}$ of $c^{\mathrm{\prime }}$,

Steps 2.1 and 3.1 show that $[a^{\mathrm{\prime }}]=[a]$ for either permitted change, so $\beta [c]$ is well defined.