The canonical map is natural

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 canonical evaluation map into the bidual, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a normed $X$, define $J_X:X\to X^{\ast \ast },(J_{X}x)(f)=f(x)(f\in X^{\ast })\mathrm{.}$ With the dual norm from def-dual-space-of-a-normed-space, evaluation is linear in $f$ and $\mathrm{\mid }(J_{X}x)(f)\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$, so $J_{X}x$ is a bounded functional on $X^{\ast }$. The map is canonical and uses no chosen basis or conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The transpose is bounded with the same norm, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. For a bounded linear $T:X\to Y$ between normed spaces, $T^{\ast }:Y^{\ast }\to X^{\ast }$ is bounded linear and $\mathrm{\parallel }{T}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$.