Baer sum of two extensions of cyclic groups

Any middle group in an extension of $\mathbb{Z}\mathrm{/}2$ by $\mathbb{Z}\mathrm{/}2$ has four elements. Transporting its law and endpoint maps to a fixed four-element set shows that the classes have a finite set realization. The extension $e$ is nonsplit because every element of $\mathbb{Z}\mathrm{/}4$ above $1\in \mathbb{Z}\mathrm{/}2$ has order four.

The diagonal pullback has middle group $D=\{(x,y)\in (\mathbb{Z}\mathrm{/}4{)}^2:x\equiv y(\mathrm{m}\mathrm{o}\mathrm{d}2)\}$ and kernel map $(a,b)\mapsto (2a,2b)$. Its codiagonal pushout is $E=(D\oplus \mathbb{Z}\mathrm{/}2)\mathrm{/}L$, where $L=\{((2a,2b),-(a+b)):a,b\in \mathbb{Z}\mathrm{/}2\}$. The endpoint maps are $j(c)=[((0,0),c)]$ and $q([((x,y),c)])=x<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$.

The map $j$ is injective since a relation with first component $(0,0)$ has $a=b=0$. If $q([((x,y),c)])=0$, write $x=2a,y=2b$; its class equals $j(c+a+b)$. Thus $\ker q=\mathrm{im}j$, and $q$ is surjective. The class $t=[((1,-1),0)]$ has $q(t)=1$ and $2t=[((2,2),0)]=0$, the relation for $a=b=1$.

Therefore $s(1)=t$ defines a homomorphic section of $q$. The map $(c,b)\mapsto j(c)+s(b)$ is an endpoint-preserving isomorphism $(\mathbb{Z}\mathrm{/}2{)}^2\to E$: surjectivity and injectivity follow from the kernel description and $qs=1$. Hence the Baer sum is split and equals the zero class.