Interesting properties in $...\to K(\mathbb{Z}_4,1) \overset{f}{\to} K(\mathbb{Z}_2,1)\overset{g}{\to}K(\mathbb{Z}_2,2) \to ...$

Question

Let $K(G,n)$ be the Eilenberg Maclane space.

Consider the map from $$ K(\mathbb{Z}_2,1) \to K(\mathbb{Z}_4,1) \overset{f}{\to} K(\mathbb{Z}_2,1)\overset{g}{\to}K(\mathbb{Z}_2,2) \to \dots, $$

It looks that we can represent the map from $K(\mathbb{Z}_2,1)\to K(\mathbb{Z}_2,2)$ relating to the generator of cohomology group $$ q \in H^2(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^2(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2. $$

Say $$ p \in H^1(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^1(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2, $$ then $q$ and $p$ are group cocycles related by $$ q = p \cup p. $$

1. Roughly, say due to the exact sequence above, we may say the kernel (in $K(\mathbb{Z}_2,1)$) of the map $g$, such that $$ q = p \cup p =0, $$ matches the image (in $K(\mathbb{Z}_2,1)$) of $$ f. $$

2. Can we see explicitly that $q = p \cup p$ may be non-zero in $H^2(K(\mathbb{Z}_2,1),\mathbb{Z}_2)=H^2(B\mathbb{Z}_2,\mathbb{Z}_2)=\mathbb{Z}_2$, but $q =0$ when we pull back via $f$ from the $K(\mathbb{Z}_2,1)$ to $K(\mathbb{Z}_4,1)$? Can this be demonstrated explicitly at the level of group cocycle?

Answer 1

Represent $p$ by the identity map $id: \mathbb{Z}_2 \to \mathbb{Z}_2$. Then $(p\cup p)(a,b) = p(a)p(b)$ is non-zero only on the 2-chain $(1,1)$. Namely, as a polynomial mod 2, $(p\cup p)(a,b) = ab$. When lifted to $\mathbb{Z}_4$, we get $p\cup p(a,b)=ab \mod 2$. Consider the function $\gamma: \mathbb{Z}_4\to \mathbb{Z}_2$ defined by $\gamma(x) = \frac{x^2-x}{2} \mod{2}$, namely $\gamma(0)=0,\gamma(1)=0,\gamma(2)=1,\gamma(3)=1$. Then $$d\gamma(x,y)=\gamma(x+y)-\gamma(x)-\gamma(y)=xy,$$ by $(x+y)^2-x^2-y^2 + (x +y)-x-y=2xy$. Note that $\gamma$ is really well defined because always $x^2 \equiv x \mod 2$ and you can see that if you add a multiple of $4$ to $x$ it don't changes the residue mod $2$ of the answer. So $d\gamma = p\cup p$ in $C^2(\mathbb{Z}_4,\mathbb{Z}_2)$.