Linear mapping/Dual mapping/Functorial properties/Fact/Proof

Proof

1. For ${}f\in {W}^{*}$, we have

   ${}{\left(\varphi \circ \psi \right)}^{*}(f)=f\circ {\left(\varphi \circ \psi \right)}={\left(f\circ \varphi \right)}\circ \psi =\varphi ^{*}(f)\circ \psi =\psi ^{*}(\varphi ^{*}(f))\,.$

2. This follows directly from ${}f\circ \operatorname {Id} _{V}=f$.
3. Let ${}f\in {V}^{*}$ and

   ${}\psi ^{*}(f)=0\,.$

   Because of the surjectivity of ${}\psi$, there exist for every ${}v\in V$ a ${}u\in U$ such that ${}\psi (u)=v$. Therefore

   ${}f(v)=f(\psi (u))=(\psi ^{*}(f))(u)=0\,,$

   and ${}f$ is itself the zero mapping. Due to fact, ${}\psi ^{*}$ injective.

4. The condition means that we may consider ${}U\subseteq V$ as a linear subspace. Because of fact, we can write

   ${}V=U\oplus U'\,$

   with another ${}K$-linear subspace ${}U'\subseteq V$. A linear form

   $g\colon U\longrightarrow K$

   can always be extended to a linear form

   ${\tilde {g}}\colon V\longrightarrow K,$

   for example, by defining ${}{\tilde {g}}$ on ${}U'$ to be the zero form. This means the surjectivity.

To fact