Let "D" be an open set containing σ("T"). Suppose Γ = {γ"i"} and Ω = {ω"j"} are two (finite) collections of Jordan curves satisfying the assumption given for the functional calculus. We wish to show  Let Ω′ be obtained from Ω by reversing the orientation of each ω"j", then  Consider the union of the two collections Γ ∪ Ω′. Both Γ ∪ Ω′ and σ("T") are compact. So there is some open set "U" containing Γ ∪ Ω′ such that σ("T") lies in the complement of "U". Any "a" in the complement of "U" has winding number "n"(Γ ∪ Ω′, "a") = 0 and the function  is holomorphic on "U". So the vector-valued version of Cauchy's theorem gives  i.e. Hence the functional calculus is well-defined. Consequently, if "f"1 and "f"2 are two holomorphic functions defined on neighborhoods "D"1 and "D"2 of σ("T") and they are equal on an open set containing σ("T"), then "f"1("T") = "f"2("T"). Moreover, even though the "D"1 may not be "D"2, the operator ("f"1 + "f"2) ("T") is well-defined. Same holds for the definition of ("f"1·"f"2)("T"). So far the full strength of this assumption has not been used. For convergence of the integral, only continuity was used. For well-definedness, we only needed "f" to be holomorphic on an open set "U" containing the contours Γ ∪ Ω′ but not necessarily σ("T"). The assumption will be applied in its entirety in showing the homomorphism property of the functional calculus. The linearity of the map "f" ↦ "f"("T") follows from the convergence of the integral and that linear operations on a Banach space are continuous. We recover the polynomial functional calculus when "f"("z") = ∑0 ≤ "i" ≤ "m"  "ai zi" is a polynomial. To prove this, it is sufficient to show, for "k" ≥ 0 and "f"("z") = "zk", it is true that "f"("T") = "Tk", i.e.  for any suitable Γ enclosing σ("T").