I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to $0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is $Sq^1$ and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by $2$ on $2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to $2H^\ast(X,\mathbb Z)$ (instead of $H^\ast(X,\mathbb Z)$) when Z)$). When each cohomology group$H^\ast(X,\mathbb Z)$is finitely generated this means concretely that you "keep" each$\mathbb Z$-factor (as well as odd torsion) and downgrade each$\mathbb Z/2^n$to$\mathbb Z/2^{n-1}$. In particular the difference between the dimension of$H^n(X,\mathbb Z/2)$and that of the$Sq^1$-cohomology is equal to the number of$\mathbb Z/2$-factors in$H^n(X,\mathbb Z)$and$H^{n+1}(X,\mathbb Z)$. I found a reference to Q2. In Madsen, Milgram: The classifying spaces for surgery and cobordism of manifolds, Ann of Math Studies 92 where they refer to Browder: Torsion in H-spaces, Ann of Math 74 for the Bockstein s.s. of$K(\mathbb Z_{(2)},n)$and$K(\mathbb Z/2,n)$. The Madsen-Milgram book also contains other examples of computations with the Bss. 3 added 380 characters in body I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to$0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is$Sq^1$and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by$2$on$2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to$2H^\ast(X,\mathbb Z)$(instead of$H^\ast(X,\mathbb Z)$) when each cohomology group$H^\ast(X,\mathbb Z)$is finitely generated this means concretely that you "keep" each$\mathbb Z$-factor (as well as odd torsion) and downgrade each$\mathbb Z/2^n$to$\mathbb Z/2^{n-1}$. In particular the difference between the dimension of$H^n(X,\mathbb Z/2)$and that of the$Sq^1$-cohomology is equal to the number of$\mathbb Z/2$-factors in$H^n(X,\mathbb Z)$and$H^{n+1}(X,\mathbb Z)$. I found a reference to Q2. In Madsen, Milgram: The classifying spaces for surgery and cobordism of manifolds, Ann of Math Studies 92 where they refer to Browder: Torsion in H-spaces, Ann of Math 74 for the Bockstein s.s. of$K(\mathbb Z_{(2)},n)$and$K(\mathbb Z/2,n)$. The Madsen-Milgram book also contains other examples of computations with the Bss. 2 added 209 characters in body I think the easiest way to understand the Bockstein spectral sequence is through the exact couple coming from the long exact sequence of cohomology associated to$0\to\mathbb Z\to\mathbb Z\to \mathbb Z/2\to0$. This shows first that indeed the first differential is$Sq^1$and tells you that the next page is the direct sum of the cokernel and kernel (shifted one step) of multiplication by$2$on$2H^\ast(X,\mathbb Z)$. Hence it is like what you would get from applying the universal coefficient formula to$2H^\ast(X,\mathbb Z)$(instead of$H^\ast(X,\mathbb Z)$) when each cohomology group$H^\ast(X,\mathbb Z)$this means concretely that you "keep" each$\mathbb Z$-factor (as well as odd torsion) and downgrade each$\mathbb Z/2^n$to$\mathbb Z/2^{n-1}$. In particular the difference between the dimension of$H^n(X,\mathbb Z/2)$and that of the$Sq^1$-cohomology is equal to the number of$\mathbb Z/2$-factors in$H^n(X,\mathbb Z)$and$H^{n+1}(X,\mathbb Z)\$