Inducing a linear map $V \to (V/(V^{\perp R}))^*$ I am not confident in working with quotient spaces, so I want to know if the argument that I used to prove the question is right and, if not, some hint and some tips to work better with quotient.
The question is: Let $B$ be a bilinear form on $V$. Set $V^{\perp L} = \{v \in V : v \perp V\}$ and $V^{\perp R} = \{v \in V : V \perp v\}$. Since $B(v,w+w') = B(v,w)$ when $w' \in V^{\perp R}$, $L_B$ $(L_B: V \to V^*,$  where $L_B(v) = B(v,-))$ induces a linear map $V \to (V/V^{\perp R})^*$. Show this linear map has kernel $V^{\perp L}$, so we get a linear embedding $V/V^{\perp L} \hookrightarrow (V/V^{\perp R})^*$.
My proof:
Ker$(L_B) = \{v \in V : L_B(v) = 0\}$. We know that $L_B(v) = B(v,-) + V^{\perp R*}$ (dual of $V^{\perp R}$), then, if $L_B(v) = 0$ we have that $B(v,-) + V^{\perp R*} = 0$. Note that $V^{\perp R*} = \{B' \in Bil(V)^1 : B'(-,v') = 0$ for all $v' \in V^{\perp R}\}$, so we have $B(v,-) + B(v,v') = 0 \implies B(v,-) = 0$, but then $B(v,-) \in V^{\perp L*}$, so $v \in V^{\perp L}$.
Does that make any sense?
1 = $Bil(V)$ is the set of all bilinear forms on $V$.
 A: I'll write $\def\rad{\mathrm{rad}}$$\rad_R(B)$ and $\rad_L(B)$ instead of $V^{\perp R}$ and $V^{\perp L}$ for the right/left radicals of $B$.
We don't quite have $L_B(v) = B(v,-) + \rad_R(B)$ since $L_B(v)$ is a linear functional, and not a class. It'd be more correct to write $L_B(v)\colon V/\rad_R(B) \to \Bbb K$, given by $$L_B(v)(w+\rad_R(B)) = B(v,w).$$This is well defined as the question indicates, since if $w-w' \in \rad_R(B)$, then $B(v,w-w')=0$ and so $B(v,w) = B(v,w')$. This says that each $L_B(v)$ (initially defined in $V$) passes to the quotient, and since $v \in V$ was arbitrary, (using $L_B$ again) we have $L_B\colon V \to (V/\rad_R(B))^\ast$.
Good. Now that we understand what is going on, this'll be a breeze. Let $v \in \ker L_B$. Then $L_B(v)\colon V/\rad_R(B)\to \Bbb K$ is the zero functional. Let $w \in V$ be arbitrary. Then we can consider the class $w + \rad_R(B)$ and apply $L_B(v)$ to it. This yields $$0=L_B(v)(w+\rad_R(B)) = B(v,w) \implies v \in \rad_L(B).$$So $\ker L_B \subseteq \rad_L(B)$.
Now assume that $v \in \rad_L(B)$. We want to see that $L_B(v)$ is the zero functional. So take a class $w + \rad_R(B)$ and compute $$L_B(v)(w+\rad_R(B)) = B(v,w) \stackrel{(\ast)}{=} 0,$$where in $(\ast)$ we used $v \in \rad_L(B)$. So $v \in \ker L_B$.
We conclude that $\ker L_B = \rad_L(B)$, and so $L_B$ induces a map $V/\rad_L(B) \to (V/\rad_R(B))^\ast$, as wanted.
