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A state with $k$ B smoothings has at most $s_A+k$ circles, so its largest A exponent is at most $c(D)+2s_A-2$. The dual argument bounds the smallest exponent below by $-c(D)-2s_B+2$. For a connected link diagram, the Turaev surface inequality bounds the resulting breadth by $4c(D)$. For a connected reduced alternating diagram, adequacy prevents cancellation at both extremes and planar Euler characteristic gives equality. A disconnected projection needs the additional term in the Jones polynomial breadth bound for a disconnected diagram.

False as written for arbitrary links. The two-component unlink has a diagram with no crossings butThe 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 ${\mathrm{br}}_A⟨D⟩\le 2n+2s_A+2s_B-4$. For a connected projection, the Turaev surface has Euler characteristic $s_A+s_B-n\le 2$, 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 $\mathrm{br}{V}_L\le n$.

For a diagram with $k$ projection components, sum the same Euler characteristic inequality over the components. Then $s_A+s_B\le n+2k$ and the Jones polynomial breadth bound for a disconnected diagram is $\mathrm{br}{V}_L\le n+k-1$. The extra $k-1$ is indispensable, as the unlink shows.