Construct a canonical isomorphism from $\text{Ker}(g^*)$ to $\text{Coker}(g)$ if $g: L \to M$, and $L,M$ are finite dimensional vector spaces. ($g^*$ is the dual map).
$\text{Coim} (g) = L/\text{Ker}(g), \text{Coker}(g) = M/\text{Im}(g)$.
If $g: L \to M$, then $g^*: M^* \to L^*$.
$\text{Ker}(g^*) = \{f \in M^*: (f o g)(l) = f(g(l)) = 0, \forall l \in L\}$
$\text{Coker}(g) = \{m + \text{Im} (g): \forall m \in M\}$
Since $f \in M^*$, then it takes $f: M \to F$, where $F$ is some field of scalars. Let $\{g(e_1), g(e_2),\dots, g(e_n)\}$ be the basis of $Im(g)$, extend it to the $\{g(e_1), g(e_2), ..., g(e_n), m_1,\dots,m_k\}$ basis of $M$. Then $f(m) = 0$ if $m \in \text{Im}(g)$ and $f(m) = a_m$ for some $a_m \in F$.
Suppose there is a canonical isomorphism $T:\text{Ker}(g^*) \to \text{Coker}(g)$, then $T(f) = m + \text{Im}(g)$ for some $f \in \text{Ker}(g^*)$. However, I don't see how can $\text{Ker}(g^*)$ be mapped to $\text{Coker}(g)$. How should one get $m$ from $f \in \text{Ker}(g^*)$?
I think it should be $\ker(g^\ast)\to\operatorname{coker}(g)^\ast$, for consider the exact sequence $$L\xrightarrow g M\xrightarrow p M/\operatorname{Im}g\to\{0\}$$ where $p$ denote the canonical projection. By dualizing, you get the following commutative diagram with exact rows$\require{AMScd}$: \begin{CD} \{0\}@>>>\ker(g^\ast)@>>> M^\ast@>g^\ast>>L^\ast\\ @.@V\sim VV@|@|\\ \{0\}@>>>(M/\operatorname{Im}g)^\ast@>>p^\ast>M^\ast@>>g^\ast>L^\ast \end{CD} Consequently, the required isomorphism $\ker(g^\ast)\to\operatorname{coker}(g)^\ast$ is the inverse of the isomorphism $\operatorname{coker}(g)^\ast\to\ker(g^\ast)$ induced by $p^\ast$. For finite dimensional vector spaces, $\operatorname{coker}(g)^\ast$ is (non-naturally) isomorphic to $\operatorname{coker}(g)$.