Show that coker$(A^*):=V^*/\text{im}(A^*)$ is canonically isomorphic to $(\ker(A))^*$ Let $K$ be a field and $V,V'$ finite dimensional $K$-vector spaces. Let $A:V\to V'$ be a homomorphism and $A^*:V'^*\to V^*$ the corresponding dual map.
How do I show that $\operatorname{coker}(A^*) := V^*/\operatorname{im}(A^*)$ is canonically isomorphic to $(\ker(A))^*$? Any help is appreciated
 A: Let $\iota : \ker A \hookrightarrow V$ be the inclusion map. Then note that $\iota^* : V^* \to (\ker A)^*$ induces a linear isomorphism $V^*/(\ker \iota^*) \to \operatorname{im} \iota^*$ that maps each $\alpha+\ker \iota^* \in V^*/(\ker \iota^*)$ to $\iota^*(\alpha)$, the restriction of $\alpha$ to $\ker A$. Now, it is easy to see the following:

*

*$\ker \iota^* = \{\alpha \in V^* : \alpha(v) = 0 \textrm{ for all } v\in\ker A\} = (\ker A)^0 = \operatorname{im} A^*$, and that

*$\operatorname{im} \iota^* = (\ker A)^*$, since any linear function $\ker A \to k$ can be extended to a linear function $V \to k$ (here $k$ is the ground field).

Added: Given two vector spaces, $V$ and $W$, and $U$ a subspace of $V$, any linear map $f : V \to W$ with $U \subseteq \ker f$ induces a linear map $\widetilde{ f\,} : V/U \to W$ given by $\widetilde{ f\,}(v+U) = f(v)$. Observe that the condition $U \subseteq \ker f$ tells us that $v_1+U=v_2+U$ implies $f(v_1) = f(v_2)$, that is, $\widetilde{ f\,}$ is well defined. Moreover, note that $\ker \widetilde{ f\,} = (\ker f)/U$ and that $\operatorname{im} \widetilde{ f\,} = \operatorname{im} f$. So, in the case that $U = \ker f$ and $W = \operatorname{im} f$, we see that $\widetilde{ f\,}$ is an isomorphism.
