= Solution
For an oriented three-manifold M with prescribed framed boundary endpoints P, the <relative Kauffman bracket skein module> is the complex vector space generated by properly embedded framed <tangles>, modulo framed <isotopy> and the local bracket relations. A crossing resolves with coefficients A and $A^{-1}$, and adjoining an unknotted <circle> multiplies by $\delta=-A^2-A^{-2}$. We now use the full bracket convention: the empty diagram has value one and a <circle> has value $\delta$. For a nonempty uncolored <link> this full bracket is $\delta$ times the reduced bracket used earlier. Fixing this normalization is essential for the <quantum dimensions>. Gluing compatible boundary pieces composes skeins; disjoint union gives <tensor products>. With no endpoints this is the <Kauffman bracket skein module>. Framing is retained, so a Reidemeister-I curl is a framing change rather than an allowed <isotopy>.
For $A\ne0$, the <Temperley-Lieb algebra> $\mathrm{TL}_n(A)$ is the skein algebra of a rectangle cylinder with n marked endpoints on each horizontal side, multiplication being vertical stacking. Resolve crossings to get a basis of crossing-free matchings of its $2n$ endpoints, with every closed <circle> removed with factor $\delta$. There are $\frac1{n+1}\binom{2n}{n}$ such matchings. Conversely multiplication of these matchings by joining endpoints defines an algebra satisfying the skein relations, which proves their independence. Writing $e_i$ for an adjacent cap-cup and 1 for the n vertical strands gives
$$
e_i^2=\delta e_i,\qquad e_ie_{i\pm1}e_i=e_i,\qquad e_ie_j=e_je_i\quad(|i-j|>1).
$$
Every crossing-free matching other than 1 contains a turnback; removing an outermost cap successively expresses it in these generators. Thus these diagram rules give the usual generator presentation as well.
Put $[m]=(A^{2m}-A^{-2m})/(A^2-A^{-2})$, interpreted as the Laurent sum $\sum_{s=0}^{m-1}A^{2(m-1-2s)}$ if its denominator vanishes. Set $\Delta_j=(-1)^j[j+1]$. These <quantum integers> satisfy $\Delta_0=1$, $\Delta_1=\delta$, and $\Delta_j=\delta\Delta_{j-1}-\Delta_{j-2}$. The <Jones-Wenzl idempotent> $f_n$ is characterized by identity-matching coefficient one and
$$
f_n^2=f_n,\qquad e_if_n=f_ne_i=0\quad(1\leq i<n).
$$
Starting with $f_0=1$, $f_1=1$, set $F=f_{n-1}\otimes1$ and $C=Fe_{n-1}F$. Whenever the denominators are nonzero the recursion is
$$
\boxed{f_n=F-\frac{\Delta_{n-2}}{\Delta_{n-1}}C.}
$$
Here is the induction proving its properties. Closing the last strand of $f_{n-1}$ gives $(\Delta_{n-1}/\Delta_{n-2})f_{n-2}$. Consequently stacking two copies of C, and straightening the middle cap-cup, gives $C^2=\mu C$, where $\mu=\Delta_{n-1}/\Delta_{n-2}$. Also $FC=CF=C$. Expanding $(F-\mu^{-1}C)^2$ proves <idempotence>. The earlier turnbacks are killed by F; the last is killed because the same cap-cup calculation gives $Ce_{n-1}=\mu Fe_{n-1}$ and its reflected identity. Only F contributes to the identity matching, with coefficient one.
For completeness, the partial-closure formula used in that induction follows from the recursion itself. Closing the last strand of F gives $\delta f_{n-1}$; closing the last strand of C straightens its cap-cup to $f_{n-1}$. Hence partial closure of $f_n$ is $(\delta-\Delta_{n-2}/\Delta_{n-1})f_{n-1}=(\Delta_n/\Delta_{n-1})f_{n-1}$. Starting at $f_1$, this proves both the inductive formula and total closure $\operatorname{tr}(f_n)=\Delta_n$. Finally, any nonidentity matching is in the turnback ideal. If two normalized projectors kill that ideal, multiplying them gives each of them, proving uniqueness. Their normalization also gives absorption of smaller projectors on consecutive substrands. At exceptional A one must check the denominators; the recursion does not assert existence of every projector there.
A positive framing twist of an n-strand projected cable consists of n individual positive curls and $n(n-1)$ positive braid crossings. Each curl contributes $-A^3$. A crossing is $A1+A^{-1}e_i$, so it acts on $f_n$ as A because the turnback vanishes. This proves the depicted <Jones-Wenzl twist eigenvalue> identity, including its exponent:
$$
\boxed{\theta_n f_n=(-1)^nA^{n(n+2)}f_n.}
$$
A negative framing twist acts by the inverse scalar.
Choose $A=e^{\pi i/(2r)}$, with an integer $r\geq3$. Then $[r]=0$ but $[1],\ldots,[r-1]$ are nonzero. The projector $f_{r-1}$ exists and has trace zero. Pass to the <reduced Temperley-Lieb skein theory> by setting it to zero, including every diagram containing it. The surviving colors are $j=0,\ldots,r-2$, with dimensions and twists
$$
d_j=\Delta_j=(-1)^j\frac{\sin((j+1)\pi/r)}{\sin(\pi/r)},\qquad\theta_j=(-1)^jA^{j(j+2)}.
$$
Their dimensions are nonzero. The recurrence decomposes $f_j\otimes1$ into the two orthogonal channels $f_{j+1}$ and $f_{j-1}$: the first is the new projector, while its complementary cap-cup <idempotent> is the latter, with inclusion/projection normalized by the nonzero partial-closure scalar. At the upper endpoint the $r-1$ channel vanishes. Iterating gives a complete decomposition of every tensor power into these colors. A matching between projected colors of different sizes has a turnback on the larger side and vanishes; on the same size only the identity matching survives. Thus their endomorphism spaces are one-dimensional and all <matrix> blocks split. Applying the recurrence to the decompositions proves the <fusion of Jones-Wenzl colors> rule
$$
a\otimes b=\bigoplus_cN_{ab}^c c,\qquad N_{ab}^c=\begin{cases}1,&a+b+c\text{ even},\ |a-b|\leq c\leq\min(a+b,2r-4-a-b),\\0,&\text{otherwise}.\end{cases}
$$
One can verify the induction by multiplying by color 1: both sides satisfy the same adjacent-color recurrence and the same zero boundary at $r-1$. Taking closures proves $d_ad_b=\sum_cN_{ab}^cd_c$. Cups and caps make every color self-dual, so these fusion coefficients are symmetric in all three labels.
We next prove the <Kirby color handle-slide identity>, rather than assume it. In the annular skein space let $\widehat f_b$ be the closure of color b and define the <Kirby color> $\Omega=\sum_bd_b\widehat f_b$. For a fixed color a, choose inclusion/projection maps $i_c:c\to b\otimes a$, $p_c:b\otimes a\to c$ for the fusion channels, with $p_ci_c=1_c$ and $\sum_ci_cp_c=1_{b\otimes a}$. Bend the a strand with a cup and cap, obtaining
$$
U_c^b=(p_c\otimes1_a)(1_b\otimes\operatorname{coev}_a):b\longrightarrow c\otimes a,\qquad V_c^b=(1_b\otimes\operatorname{ev}_a)(i_c\otimes1_a):c\otimes a\longrightarrow b.
$$
Closing the resulting diagrams and straightening the bent strand gives $\operatorname{tr}(V_c^bU_c^b)=\operatorname{tr}(p_ci_c)=d_c$. Since b is simple, $V_c^bU_c^b=(d_c/d_b)1_b$. The symmetric fusion rule supplies all channels of $c\otimes a$, so their normalized projections are complete. Therefore
$$
\sum_b\frac{d_b}{d_c}U_c^bV_c^b=1_{c\otimes a},\qquad\boxed{\sum_bd_bU_c^bV_c^b=d_c1_{c\otimes a}.}
$$
To slide an a-colored strand over an $\Omega$-colored component, cut at the slide band and insert the resolution $\sum_ci_cp_c$. Carrying the band across that loop bends its two maps into U and V. In each outgoing channel c, the boxed identity replaces the sum over the old loop color b by exactly the new loop weight $d_c$; the rest of the diagrams are isotopic as framed ribbons. This proves invariance for fixed a. Summing a with its own weight proves invariance when both surgery components are colored by $\Omega$. The band framing is carried in this argument, as required for a framed <handle slide>.
We must also show that the stabilization normalization is nonzero. Let $H_{ij}$ be the full bracket of the zero-framed <Hopf link> with colors i,j. On a fusion channel k, its double braiding has eigenvalue $\theta_k/(\theta_i\theta_j)$: the full twist of the combined ribbon equals double braiding followed by the two separate twists. Taking the trace gives the balancing identity
$$
H_{ij}=\frac1{\theta_i\theta_j}\sum_kN_{ij}^kd_k\theta_k.
$$
For $j=1$ there are only the $i-1,i+1$ channels; substitution of their dimensions and twists gives $H_{i1}/d_i=-(A^{2i+2}+A^{-2i-2})$, with a vanished endpoint channel simply omitted. A color-1 meridian therefore acts on the simple color i by that scalar. The annular closures obey $\widehat f_j=X\widehat f_{j-1}-\widehat f_{j-2}$, where X is a color-1 core, by the two-channel fusion decomposition. Applying this recurrence to meridians gives the <Hopf pairing of Jones-Wenzl colors>
$$
H_{ij}=(-1)^{i+j}[(i+1)(j+1)].
$$
Indeed the recurrence is the sine identity $\sin((j+1)x)=2\cos x\sin(jx)-\sin((j-1)x)$ with $x=(i+1)\pi/r$, and its initial values are $H_{i0}=d_i$ and the computed $H_{i1}$. Finite sine orthogonality now gives
$$
H^2=\mathcal D^2I,\qquad\mathcal D^2=\sum_jd_j^2=\frac{r}{2\sin^2(\pi/r)},\qquad\mathcal D>0.
$$
For example, orthogonality follows by writing products of sines as differences of cosines and summing the finite geometric series of rth roots of unity; the result is $\sum_{k=1}^{r-1}\sin(ik\pi/r)\sin(jk\pi/r)=\frac r2\delta_{ij}$. Thus no degeneracy remains in this pairing.
The <Gauss sums of Jones-Wenzl colors> are $p_\pm=\sum_jd_j^2\theta_j^{\pm1}$, the evaluations of the plus- and minus-one-framed $\Omega$-colored unknots. Balancing and dimension fusion prove
$$
\sum_id_i\theta_iH_{ij}=\theta_j^{-1}\sum_{i,k}d_iN_{ij}^kd_k\theta_k=p_+\theta_j^{-1}d_j.
$$
In the last step $\sum_id_iN_{ij}^k=d_jd_k$, by symmetry of the fusion coefficients. In vector notation this is $HTd=p_+T^{-1}d$, with $T=\operatorname{diag}(\theta_j)$. The <matrix> $H$ and <vector> $d$ are real, and each twist has modulus one, so conjugation gives $HT^{-1}d=p_-Td$. Apply H again and use its orthogonality to obtain $\boxed{p_+p_-=\mathcal D^2}$. Hence neither factor vanishes.
We may use the surgery existence theorem and <Kirby calculus>: every closed oriented three-manifold has an integral framed-link surgery description, and two descriptions are related by <handle slides> and insertion/removal of disjoint plus- or minus-one-framed unknots. For an m-component surgery <link> L let $\sigma(L)$ be the <signature> of its symmetric <surgery linking matrix>, put $\kappa=p_+/\mathcal D$, and define the <Jones polynomial surgery invariant>
$$
\boxed{\tau_r(M_L)=\mathcal D^{-m}\kappa^{-\sigma(L)}\langle L(\Omega,\ldots,\Omega)\rangle.}
$$
A <handle slide> preserves the colored bracket by the proved identity and changes the linking <matrix> by congruence, preserving its <signature>. Adding a disjoint plus-one-framed <unknot> multiplies the bracket by $p_+$ and changes $(m,\sigma)$ to $(m+1,\sigma+1)$, so the net factor is $p_+/(\mathcal D\kappa)=1$. For a minus-one-framed <unknot> it is $\kappa p_-/\mathcal D=p_+p_-/\mathcal D^2=1$. <Kirby calculus> therefore proves that the displayed value is a well-defined oriented three-manifold invariant. Our normalization has $\tau_r(S^3)=1$; reversing <orientation> conjugates the value. Its construction uses precisely the bracket and colored projectors underlying the <Jones polynomial>.
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