The dual of a quotient is its annihilator

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The dual space X^* of a normed space and its dual norm, with its stated hypotheses: Let $X$ be a normed space over the scalar field $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ in the literal definition and $\mathbb{K}=\mathbb{C}$ by the convention of rem-real-and-complex-normed-space-convention. The dual space of $X$ is $X^{\ast }:=\mathcal{B}(X,\mathbb{K}),$ the space of bounded linear functionals on $X$ (def-space-of-bounded-linear-operators). Each $f\in X^{\ast }$ is in particular a linear functional in the algebraic sense, so $X^{\ast }$ is a subspace of the algebraic dual from def-algebraic-dual-and-linear-functional. The dual norm on $X^{\ast }$ is the operator norm: $\mathrm{\parallel }f{\mathrm{\parallel }}_{X^{\ast }}:=\mathrm{\parallel }f\mathrm{\parallel }=\sup \{\mathrm{\mid }f(x)\mathrm{\mid }:\mathrm{\parallel }x\mathrm{\parallel }\le 1\}\mathrm{.}$

From Annihilator notation and the preannihilator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$ and arbitrary subsets $M\subseteq X$, $N\subseteq X^{\ast }$, define $M^{\perp }=\{f\in X^{\ast }:f(m)=0\text{ for all }m\in M\},{}^{\perp }N=\{x\in X:f(x)=0\text{ for all }f\in N\}\mathrm{.}$ Here $X^{\ast }$ is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on $\mathrm{span}M$, since linearity makes vanishing on $M$ equivalent to vanishing on its span. The preannihilator lies in $X$, not in $X^{\ast \ast }$. Empty sets impose no conditions: ${\mathrm{\varnothing }}^{\perp }=X^{\ast }$ and ${}^{\perp }\mathrm{\varnothing }=X$.

From A bounded operator that vanishes on a subspace factors uniquely through the normed quotient, with its stated hypotheses: Let $X$ and $Y$ be normed spaces over the same scalar field, let $M\le X$ be a closed linear subspace, let $q:X\to X\mathrm{/}M$ be the quotient map, and let $T:X\to Y$ be a bounded linear operator with $M\subseteq \ker T$. Then there is a unique bounded linear operator $\bar{T}:X\mathrm{/}M\to Y$ such that $\bar{T}\circ q=T,$ and moreover $\mathrm{\parallel }\bar{T}\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$.