 ADHM factorization identity

With $z_1=x^1+i{x}^2$, $z_2=x^3+i{x}^4$, formThe two ADHM construction constraints say that ${\mathcal{D}}_z{\mathcal{D}}_z^{†}$ has zero off-diagonal blocks and equal diagonal blocks, so it equals $1_2\otimes f^{-1}$. Assume it is invertible everywhere. Choose an orthonormal kernel bundle frame $\Psi$, so ${\mathcal{D}}_z\Psi =0$ and ${\Psi }^{†}\Psi =1_N$, and set $A={\Psi }^{†}d\Psi$. The orthogonal complement projector is $1-\Psi {\Psi }^{†}={\mathcal{D}}_z^{†}(1_2\otimes f){\mathcal{D}}_z$. Differentiating the kernel equation givesIts space-time differential two-form entries are combinations of $d{z}_1\wedge d{\bar{z}}_1-d{z}_2\wedge d{\bar{z}}_2$, $d{z}_1\wedge d{\bar{z}}_2$, and $d{\bar{z}}_1\wedge d{z}_2$, all anti-self-dual. Thus the constructed gauge curvature satisfies the Anti-self-dual Yang-Mills equations.