The Hall square of the simple finite-field vector space

Use the small model with objects $k^n$ and all matrices between them; every finite-dimensional vector space is isomorphic to one of these. The zero object and finite direct sums exist, and each Hom set is an abelian group under pointwise addition with bilinear composition. For a linear map $f:V\to W$, its kernel and cokernel $W\mathrm{/}\mathrm{im}f$ are finite-dimensional. Choose bases of these two spaces for this single map and transport them to standard $k^d$ objects; the transported inclusion and projection retain the universal properties, so the model has kernels and cokernels. The canonical map $V\mathrm{/}\ker f\to \mathrm{im}f$ is injective because $f(v)=0$ implies $v\in \ker f$, and surjective by the definition of image, hence is an isomorphism. Thus [F1] makes the model abelian. Hom sets are finite sets of matrices over $k$. Any short exact sequence $0\to U\stackrel{i}{\to }E\stackrel{p}{\to }Q\to 0$ splits: lift a finite basis of $Q$ to $E$ and extend linearly to a section $s:Q\to E$. The map $U\oplus Q\to E$, $(u,q)\mapsto i(u)+s(q)$, is injective by applying $p$ and then using monicity of $i$; it is surjective because $e-s(p(e))$ lies in $\ker p=\mathrm{im}i$. It is an isomorphism fixing the endpoints, so every Yoneda extension is equivalent to the split one and its Yoneda extension set has one class. Therefore the model satisfies [F2].

If $g_{S,S}^L\mathrm{\ne }0$, an exact sequence $0\to S\to L\to S\to 0$ gives ${\dim }_{k}L=2$, so the only possible isomorphism class is $k^2$. Every one-dimensional subspace of $k^2$ is counted by [F3], and its quotient is one-dimensional. The nonzero vectors of $k^2$ are partitioned by their spans, each span contains $q-1$ nonzero vectors, and $k^2$ has $q^2-1$ nonzero vectors. Hence the number of lines is $(q^2-1)\mathrm{/}(q-1)=q+1$.

There are no other middle-object classes by step 1.2, and the coefficient of $[k^2]$ is the line count from step 1.2. This proves the displayed untwisted product.