Relating a 2-cocycle to specific section of a central extension

Question

I'm trying to understand from this answer how $\chi: \Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z \to \mathbb{Z}/p\mathbb{Z}$ (defined as $\chi(x, y) = 0 \ \forall x + y < p$ and $\chi(x, y) = 1 \ \forall x + y \geq p$) is the valid 2-cocycle corresponding to the (non-split) extension $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$.

As far as I know, if $\gamma$ is the section $\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}$ defined by $ x \text{ mod } p\mapsto x \text{ mod }p^2$ for $0\leqslant x<p$ then we need

$$\mathbf{\gamma(x) +_{p^2} \gamma(y) = \gamma(x +_{p} y) +_{p^2} \chi(x, y)}$$

where $\chi$ is basically a group law "correction factor" (cf. this).

Say we set $p = 5$ and pick $x = 3 \bmod 5$ and $y = 4 \bmod 5$ then

LHS = $\gamma(3 \bmod 5) +_{25} \gamma(4 \bmod 5) = (3 \bmod 25) +_{25} (4 \bmod 25) = \color{red}{7 \bmod 25}$

RHS = $\gamma (3 \bmod 5 +_5 4 \bmod 5) \ +_{25} \ \chi(3, 4) = \gamma(2 \bmod 5) +_{25} 1 \bmod 5 = 2 \bmod 25 \ +_{25} \ 1 \bmod 5 \\= 2 \bmod 25 \ +_{25} \ 1 \bmod 25 = \color{red}{3 \bmod 25}$

So clearly, LHS is not equal to RHS and thus the correction factor $\chi$ is incorrect.

Could someone please explain where I'm going wrong?

Answer 1

This is because if you write the exact sequence $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$ as $$0\to A\to B\to C\to 0$$ then you have to remember that $\chi$ is a function $C\to A$. So when $\chi(x,y)=1\in C=\mathbb{Z}/p\mathbb{Z}$, you actually have $\chi(x,y)=p\text{ mod } p^2\in B$, because the natural inclusion $A\to B$ is given by $n \text{ mod }p\mapsto pn \text{ mod }p^2$, not $n \text{ mod }p\mapsto n \text{ mod }p^2$ (which is not a group morphism).

In your example, $\chi(3,4)=1\text{ mod }5 = 5\text{ mod }25$, and you get the correct term.