The braid group <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" altimg="si1.svg"><mml:msub><mml:mrow><mml:mi>B</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi><mml:mo>,</mml:mo><mml:mi>m</mml:mi></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi mathvariant="double-struck">R</mml:mi><mml:msup><mml:mrow><mml:mi>P</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msup><mml:mo stretchy="false">)</mml:mo></mml:math> and the splitting problem of the generalised …

Let $n,m\in \mathbb{N}$, and let $B_{n,m}(\mathbb{R}P^2)$ be the set of $(n + m)$-braids projective plane whose associated permutation lies in subgroup $S_n\times S_m$ symmetric group $S_{n+m}$. We study splitting problem following generalisation Fadell-Neuwirth short exact sequence: $$1\rightarrow B_m(\mathbb{R}P^2 \setminus \{x_1,\dots,x_n\})\rightarrow B_{n,m}(\mathbb{R}P^2)\xrightarrow{\bar{q}} B_n(\mathbb{R}P^2)\rightarrow 1,$$ where map $\bar{q}$ can considered geometrically as epimorphism that forgets last $m$ strands, well existence a section corresponding fibration $q:F_{n+m}(\mathbb{R}P^2)/S_n\times S_m\to F_{n}(\mathbb{R}P^2)/S_n$, we denote by $F_n(\mathbb{R}P^2)$ $n^{th}$ ordered configuration space $\mathbb{R}P^2$. Our main results are following: if $n=1$ homomorphism $q$ admits no section, while $n=2$, then admit section. For $n\geq 3$, show $m\equiv 0, (n-1)^2\ \textrm{mod}\ n(n-1)$. Moreover, using geometric constructions, for $m=kn(2n-1)(2n-2)$, $ k\geq1$, $m=2n(n-1)$. In addition, $m\geq3$, $B_m(\mathbb{R}P^2\setminus\{x_1,\dots,x_n\})$ is not residually nilpotent $m\geq 5$, it solvable.