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?



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


