## Show a given isomorphism between a quotient space and a restriction of the dual exist

(b) Suppose that $M$ is a subspace of a vector space $V$. Corresponding to every coset $y+M^0$ of $M^0$ in $V^{\prime}$ (i.e., an element $H$ of $V^{\prime}/M^0$), there is a linear functional $z$ on $M$ (i.e., an element $z$ of $M^{\prime}$); the linear functional $z$ is defined by $z(x)=y(x)$. Prove that $z$ is unambiguously determined by the coset $H$ (that is, it does not depend on a particular choice of $y$), and that the correspondence $H \to z$ is an isomorphism between $V^{\prime}/M^0$ and $M^{\prime}$.

So define, $T: y + M^0 \to z$.

Check well defined. If $y_1+M^0=y_2+M^0$ then, $(y_1-y_2) \in M^0$. So $(y_1(x)-y_2(x))=0$ (for all $x \in M$) i.e. $y_1(x)=y_2(x)$. Now use $z(x)=y(x)$ to get $z_1(x)=z_2(x)$. So the map is well defined.

Check linearity. Since,

\begin{equation*}\begin{split}T((\alpha y_1+M^0)+(y_2+M^0)) &= (\alpha z_1+z_2) \\ &= \alpha (z_1)+(z_2)\\ &= \alpha T(y_1+M^0)+T(y_2+M^0), \end{split}\end{equation*} the transformation is linear.

To show injectivity assume that $y_1 + M^0 \neq y_2 + M^0$. So $y_1(x) \neq y_2(x)$ which means, by $z(x)=y(x)$, that $z_1(x) \neq z_2(x)$ so the map is injective.

I want to show surjectivity in the following way; show $\dim V^{\prime}/M^0 = \dim M^{\prime}$, fix a basis for each, then define the map as sending a basis vector to a basis vector.

Let $\dim V = n$, and $\dim M = m$. So $\dim V^{\prime} = n$ and $\dim M^0 = n-m$ which implies that $\dim V^{\prime}/M^0 = n-(n-m)=m=\dim M^{\prime}$. So then we map $x_i+M^0 \to x_i^*$ (where $\{x_i+M^0\}_1^{m}$ is a basis for $V^{\prime}/M^0$, and $\{x_i^*\}_1^m$ is a basis for $M^{\prime}$). Since every element in the domain and the codomain is uniquely determined by the basis for that space, and the map is one to one, the map should also be surjective.

No? This seems like magic almost so I think it is wrong.