= Solution
\b[False as written for arbitrary links.] The two-component <unlink> has a diagram with no crossings but
$$
\boxed{V_{U\sqcup U}(t)=-(t^{1/2}+t^{-1/2}),\qquad\operatorname{br}V_{U\sqcup U}=1>0.}
$$
The correct connected-diagram statement does hold, in particular for every <knot diagram>. If $s_A,s_B$ count the all-$A$ and all-$B$ <bracket smoothing state> <circles>, changing $j$ smoothings increases the circle count by at most $j$. The maximum and minimum $A$ exponents therefore lie between $n+2s_A-2$ and $-n-2s_B+2$. This gives the <Kauffman bracket breadth bound> $\operatorname{br}_A\langle D\rangle\leq2n+2s_A+2s_B-4$. For a connected projection, the <Turaev surface> has <Euler characteristic> $s_A+s_B-n\leq2$, so the bound is $4n$. <Writhe of a link diagram> normalization changes only a <monomial>, and $t=A^{-4}$ divides the breadth by four, giving $\operatorname{br}V_L\leq n$.
For a diagram with $k$ projection components, sum the same <Euler characteristic> inequality over the components. Then $s_A+s_B\leq n+2k$ and the <Jones polynomial breadth bound for a disconnected diagram> is $\operatorname{br}V_L\leq n+k-1$. The extra $k-1$ is indispensable, as the <unlink> shows.
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