Holomorphic functional calculus homomorphism

Linearity: for a common admissible cycle $\mathrm{\Gamma }$ in $U_f\cap U_g$ one has $(\alpha f+\beta g)(a)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}(\alpha f(z)+\beta g(z))R(z,a)dz=\alpha f(a)+\beta g(a)$, because the chain integral is $\mathbb{C}$-linear in the integrand.

Unit law, cycle choice: choose $r>\mathrm{\parallel }a\mathrm{\parallel }$ and apply [L5] to the compact closed disc $K=\{z:\mathrm{\mid }z\mathrm{\mid }\le r\}$ inside $\mathbb{C}$. It gives a finite polygonal cycle ${\mathrm{\Gamma }}_r$ whose trace lies outside $K$ and whose index is $1$ on $K$. Since ${\sigma }_A(a)\subseteq K$ and the constant function $1$ is entire, ${\mathrm{\Gamma }}_r$ is admissible for its calculus value; by contour independence [L1], $1(a)$ may be computed on ${\mathrm{\Gamma }}_r$.

Coordinate identity: for every admissible cycle $\mathrm{\Gamma }$ and the function $\mathrm{i}\mathrm{d}(z)=z$ one has the pointwise identity $zR(z,a)=1+aR(z,a)$ on ${\mathrm{\Gamma }}^{\ast }$, hence $\mathrm{i}\mathrm{d}(a)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}dz+a\cdot \frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}R(z,a)dz=0+a1(a)$ by [L6] and [L1].

Nested cycles: apply [L5] to the compact set $K:={\sigma }_A(a)$ and the open set $U_f\cap U_g$; this produces cycles $\beta ,\gamma$ with disjoint traces in $(U_f\cap U_g)\setminus {\sigma }_A(a)$, both admissible for $f$ and for $g$, with $n(\gamma ,w)=1$ for every $w\in {\beta }^{\ast }$ and $n(\beta ,z)=0$ for every $z\in {\gamma }^{\ast }$.

First Cauchy integral: for each fixed $w\in {\beta }^{\ast }$ the scalar function $g$ is holomorphic on $U_g$, and $\gamma$ is a cycle with trace in $U_g$ that is null-homologous in $U_g$ because its index vanishes outside $U_g$ by admissibility; [L3] gives $\frac{1}{2\pi i}{\int }_{\gamma }\frac{g(z)}{z-w}dz=n(\gamma ,w)g(w)=g(w)$.

Second Cauchy integral: for each fixed $z\in {\gamma }^{\ast }$ the function $w\mapsto \frac{f(w)}{z-w}$ is holomorphic on $U_f\setminus \{z\}$, a neighbourhood of ${\beta }^{\ast }$; and $\beta$ is null-homologous in $U_f\setminus \{z\}$, because $n(\beta ,p)=0$ for every $p\mathrm{\notin }{U}_f$ by admissibility and $n(\beta ,z)=0$ by the nesting; hence [L4] gives $\frac{1}{2\pi i}{\int }_{\beta }\frac{f(w)}{z-w}dw=0$.

Unit law, value: the compact trace of ${\mathrm{\Gamma }}_r$ lies in the open set $\{\mathrm{\mid }z\mathrm{\mid }>r\}$, so $q:={\max }_{z\in {\mathrm{\Gamma }}_r^{\ast }}\mathrm{\parallel }a\mathrm{\parallel }\mathrm{/}\mathrm{\mid }z\mathrm{\mid }<1$. Hence $R(z,a)=\frac{1}{z}(1-a\mathrm{/}z{)}^{-1}={\sum }_{n\ge 0}{a}^n{z}^{-n-1}$ uniformly on the trace. Integrating termwise by [L7], [L6] gives ${\int }_{{\mathrm{\Gamma }}_r}{z}^{-1}dz=2\pi in({\mathrm{\Gamma }}_r,0)=2\pi i$ and ${\int }_{{\mathrm{\Gamma }}_r}{z}^{-n-1}dz=0$ for $n\ge 1$. Thus $\frac{1}{2\pi i}{\int }_{{\mathrm{\Gamma }}_r}R(z,a)dz=1$ and $1(a)=1$.

The double integral: the function $H(w,z):=f(w)g(z)R(w,a)R(z,a)$ is continuous on the compact product ${\beta }^{\ast }\times {\gamma }^{\ast }$; the two-dimensional tagged Riemann sums of $H$ over refined partitions of $\beta$ and $\gamma$ converge in $A$, by the uniform-continuity mesh estimate underlying the Banach-valued contour integral in [L1] applied in both variables, so the two iterated integrals $\frac{1}{2\pi i}{\int }_{\beta }(\frac{1}{2\pi i}{\int }_{\gamma }Hdz)dw$ and $\frac{1}{2\pi i}{\int }_{\gamma }(\frac{1}{2\pi i}{\int }_{\beta }Hdw)dz$ exist and agree.

Coordinate law, value: combining [step 1.3] with [step 2.3] gives $\mathrm{i}\mathrm{d}(a)=a\cdot 1=a$.

Splitting the double integral: by the resolvent identity [L2], $R(w,a)R(z,a)=(R(w,a)-R(z,a))\mathrm{/}(z-w)$ for $w\in {\beta }^{\ast }$, $z\in {\gamma }^{\ast }$, so the double integral of [step 3.1] splits into the sum of the iterated integrals of $f(w)g(z)R(w,a)\mathrm{/}(z-w)$ and of $-f(w)g(z)R(z,a)\mathrm{/}(z-w)$; the first inner integral over $\gamma$ equals $g(w)R(w,a)$ by [step 2.1], and the second inner integral over $\beta$ equals $0$ by [step 2.2].

Multiplicativity: using [step 4.1], $f(a)g(a)=\frac{1}{2\pi i}{\int }_{\beta }f(w)R(w,a)\left(\frac{1}{2\pi i}{\int }_{\gamma }\frac{g(z)}{z-w}dz\right)dw-\frac{1}{2\pi i}{\int }_{\gamma }g(z)R(z,a)\left(\frac{1}{2\pi i}{\int }_{\beta }\frac{f(w)}{z-w}dw\right)dz=\frac{1}{2\pi i}{\int }_{\beta }f(w)g(w)R(w,a)dw=(fg)(a)$; here each resolvent factor commutes with the scalar coefficient in front of it. This is claim 2.

Inverse compatibility: for $h$ holomorphic and nowhere zero on $U_h\supseteq {\sigma }_A(a)$ the reciprocal $1\mathrm{/}h$ is holomorphic on $U_h$ and $h\cdot (1\mathrm{/}h)=1$ there; by [step 5.1] and [step 2.3], $h(a)(1\mathrm{/}h)(a)=1(a)=1$ and symmetrically $(1\mathrm{/}h)(a)h(a)=1$, so $h(a)$ is invertible with $h(a{)}^{-1}=(1\mathrm{/}h)(a)$; this is claim 5.

Polynomials and conclusion: a constant function $z\mapsto c$ is $c\cdot 1$, so its calculus value is $c1(a)=c1$ by [step 1.1] and [step 2.3]; the coordinate function has value $a$ by [step 3.2]; multiplicativity [step 5.1], linearity [step 1.1] and induction on the degree therefore assemble $p(a)={\sum }_{k=0}^n{c}_k{a}^k$ for every polynomial, which is claim 4. Claims 1, 2, 3 and 5 were proved in [step 1.1], [step 5.1], [step 2.3], [step 3.2] and [step 6.1].