proof that the Jones polynomial is an invariant

Question

On page 153 of Colin Adam's knot theory book he describes the invariance of the X polynomial (a precursor to the Jones polynomial) under the first Reidemeister move (R1). In this process Adam's seems to assume that if $L$ is a link and $L'$ is the same link after performing R1, then $\langle L' \rangle = A^{-3}\langle L \rangle$, where the $A$ mentioned is to be replaced later by $t^{\frac{1}{4}}$ as in the original Jones polynomial. Does anyone have any insights?

Apologies for the lack of clarity, will try to amend

Answer 1

He doesn't assume it : it follows from a computation based on the construction of the Jones polynomial stating that $$ V(O)=1,\qquad V(L\sqcup O)=-(A^2-A^{-2})V(L), \qquad V\big(\color{blue}{\large\times}\big)=A^{-1}V\big(\,\color{blue}{\large)\;(}\,\big) +AV\big(\color{blue}{\large\asymp}\big).$$ $O$ is the trivial knot, $L$ is any link, $L\sqcup O$ denotes a link and a disjoint trivial knot and blue symbols are to thought of as parts of link diagram. So if you compute $V$ with the links $L$ and $L'$ you get $$V(L)=A^{-1}V(L'\sqcup O)+AV(L')=-A^{-1}(A^2-A^{-2})V(L')+AV(L')=A^{-3}V(L').$$

Note. I will try to improve my answer with some diagrams.