A linear representation need not be faithful

The endpoint permutation. Since the transpositions $s_i=(ii+1)\in S_n$ satisfy $s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}$ and $s_i{s}_j=s_j{s}_i$ for $\mathrm{\mid }i-j\mathrm{\mid }>1$, the assignment $\pi :B_n\to S_n$, $\pi ({\sigma }_i)=s_i$, respects the Artin presentation and is a homomorphism. Composing it with the standard permutation representation $S_n\to {\mathrm{G}\mathrm{L}}_n(F)$ gives a linear representation ${\pi }_F:B_n⟶{\mathrm{G}\mathrm{L}}_n(F)$ of degree $n$ over any field.

The kernel is nontrivial. The exponent-sum map $\epsilon :B_n\to \mathbb{Z}$, $\epsilon ({\sigma }_i)=1$, is well defined because every Artin relator has equal total exponent on both sides; hence $\epsilon ({\sigma }_1^2)=2\mathrm{\ne }0$ and ${\sigma }_1^2\mathrm{\ne }1$ in $B_n$. The element ${\sigma }_1^2$ is pure: its image under $\pi$ is $s_1^2=1$. Therefore ${\sigma }_1^2$ is a nontrivial element of $\ker \pi$, and a fortiori of $\ker {\pi }_F$; for $n\ge 2$ the kernel of the permutation representation is the nontrivial pure braid group.

Linear does not mean faithful. Thus for every $n\ge 2$ the group $B_n$ admits the linear representation ${\pi }_F$ of degree $n$ with $\ker {\pi }_F\mathrm{\ne }1$. Linearity of $B_n$ is therefore not witnessed by an arbitrary representation: the content of Every classical braid group is linear lies in the faithfulness of ${\rho }_{\mathrm{L}\mathrm{K}\mathrm{B}}$, not merely in the existence of a representation, and the claim stated above is refuted.