22 - Isomorphism of Quotient Space Dual

Isomorphism of Quotient Space Dual

Let $U\subseteq V$ be a subspace
Then there exists a normal isomorphism

$$U^0\simeq (V/U{)}^{\prime }$$

given by $f\mapsto \overline{f}$, where $\overline{f}(v+U):=f(v)$ for $v\in V$

Proof

Let $f\in U^0$
Note that $\overline{f}$ is well-defined because $f{\mid }_U=0$

Map is linear as

$$\overline{\lambda f+h}=\lambda \overline{f}+\overline{h}\text{ for all }f,h\in U^0\text{ and }\lambda \in \mathbf{F}$$

Map is injective as

$$\overline{f}=0⟹\overline{f}(v+U)=f(v)=0\text{ for all }v\in V⟹f=0$$

Note that for the finite dimensional case, considering dimensions of both sides gives result

In general, we can construct an inverse for $f\mapsto \overline{f}$ as follows.

Let $g\in (V/U{)}^{\prime }$ then define $\hat{g}\in V^{\prime }$ by $\hat{g}(v):=g(v+U)$, then $\hat{g}$ is linear in $v$
As the map $g\mapsto \hat{g}$ is linear in $g$, with the image in $U^0$
Finally then $\hat{\overline{f}}=f$ and $\overline{\hat{g}}=g$