Disk-supported Pauli centralizers are stabilizers

Commutation of $P(a,b)$ with every $A_v$ gives $b\cdot {\mathrm{\partial }}_1^{T}v=0$ for every vertex basis vector $v$, hence ${\mathrm{\partial }}_{1}b=0$. Commutation with every $B_f$ gives $a\cdot {\mathrm{\partial }}_{2}f=0$ for every face basis vector $f$, hence ${\mathrm{\partial }}_2^{T}a=0$.

The image of ${\mathrm{\partial }}_1$ is the even-parity subspace of $C_0$: each edge boundary has even parity, while a path from a fixed vertex $v_0$ to any $v\mathrm{\ne }{v}_0$ has boundary $v+v_0$, and these vectors span that subspace. Thus $\mathrm{rank}{\mathrm{\partial }}_1=\mathrm{\mid }V\mathrm{\mid }-1$. A 2-chain lies in $\ker {\mathrm{\partial }}_2$ exactly when its coefficients agree across every adjacent pair of faces; dual-graph connectedness makes these coefficients constant, so $\dim \ker {\mathrm{\partial }}_2=1$ and $\mathrm{rank}{\mathrm{\partial }}_2=\mathrm{\mid }F\mathrm{\mid }-1$. Rank-nullity gives $\dim H_1(K_L;{\mathbb{F}}_2)=\mathrm{\mid }E\mathrm{\mid }-(\mathrm{\mid }V\mathrm{\mid }-1)-(\mathrm{\mid }F\mathrm{\mid }-1)=2$. The mod-two intersection pairing with a closed dual cycle vanishes on every face boundary, since each face boundary meets that closed dual cycle an even number of times; it is therefore well defined on $H_1$. The matrix $I(p_i,q_j)={\delta }_{ij}$ proves that $p_1,p_2$ are independent; by the dimension calculation they form a basis, and $q_1,q_2$ detect the two coefficients of every homology class.

Since $b$ is a cycle and its support is disjoint from $q_1,q_2$, its two intersection numbers vanish. The intersection matrix and the two-dimensionality in step 1.2 imply that $b$ has zero homology class; by the definition of $H_1=\ker {\mathrm{\partial }}_1\mathrm{/}\mathrm{im}{\mathrm{\partial }}_2$, there is a face chain $w$ with $b={\mathrm{\partial }}_{2}w$.

The cochain $a$ vanishes on $p_1,p_2$ because its support is disjoint from them. It also vanishes on every boundary ${\mathrm{\partial }}_2{w}^{\mathrm{\prime }}$ because ${\mathrm{\partial }}_2^{T}a=0$. Since $p_1,p_2$ represent a basis of $H_1$, every 1-cycle is a sum of their representatives and a boundary; hence $a$ vanishes on every 1-cycle. Define $\varphi$ on $\mathrm{im}{\mathrm{\partial }}_1$ by $\varphi ({\mathrm{\partial }}_{1}c)=a\cdot c$. This is well defined because two chains with the same boundary differ by a cycle, on which $a$ vanishes. Extend $\varphi$ linearly to $C_0$ by extending a basis of the finite-dimensional subspace $\mathrm{im}{\mathrm{\partial }}_1$. Then $a={\mathrm{\partial }}_1^T\varphi$, so take $u=\varphi$.

By $a={\mathrm{\partial }}_1^{T}u$ and $b={\mathrm{\partial }}_{2}w$, $X(a)$ is the product of $A_v$ over vertices with $u_v=1$, and $Z(b)$ is the product of $B_f$ over faces with $w_f=1$. Therefore $P(a,b)=X(a)Z(b)$ is the product of those vertex and face stabilizers.