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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.

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.

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 $H^{n+1}(X,\mathbb Z)$

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.

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 $\mathbb Z/2^{n-1}$.

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 $H^{n+1}(X,\mathbb Z)$