Proving $\operatorname{coker}(f^*) \cong (\ker f)^*$ for a linear map $f$

Question

Let $f: V \to W$ be linear and $V, W$ be vector spaces of finite dimension. I want to show that the cokernel, defined by $\operatorname{coker}(f^*) := V^* / \operatorname{im}(f^*)$, is isomorphic to $(\ker f)^*$.

I already tried to use the rank and nullity theorem

$$\dim \operatorname{coker} f^* = \dim V^* - \dim \operatorname{im}f^* \iff \dim V^* = \dim \operatorname{coker} f^*+\dim \operatorname{im}f^*$$

and maybe one could show $\dim \operatorname{im} f^* = \dim \operatorname{im}f$ and thus

$$\dim \operatorname{coker} f^* = \dim \ker f = \dim (\ker f)^*.$$ But how can I prove that the dimensions of the images of $f$ and $f^*$ and the dimensions of $\ker f$ and $(\ker f)^*$ are equal? Is this anyway the right way to prove it?

Answer 1

Try to take a look at the restriction map $V^*\to (\operatorname{ker} f)^*$, and show that it factors to a map $\operatorname{coker}(f^*)\to (\operatorname{ker} f)^*$.

Then you can use dimension arguments to show that it is an isomorphism.

Answer 2

Define $\Phi: V^*\to (\ker f)^*$ by $\Phi(x^*)=x^*|_{\ker f}$ for all $x^*\in V^*$. It is easy to see that $\Phi$ is a surjective linear map. Now if $x^*= f^*(y^*)$ for some $y^*\in W^*$ then $x^*|_{\ker f}(x)=y^*(f(x))=y^*(0)=0$ for all $x\in \ker f$.

Now, suppose $x^*|_{\ker f}(x)=0$ for all $x\in\ker f$. Let $\mathcal{B}$ be a basis of $\text{im}(f)$, and $\mathcal{B}'$ be a basis of $W$ containing $\mathcal{B}$. Define $y^*(y)=x^*(x)$ if $y\in \mathcal{B}$ and $y=f(x)$, and $y^*(y)=0$ if $y\in\mathcal{B}'\setminus\mathcal{B}$. Then, extend $y^*$ to the whole $W$ by linearity. It follows that $y^*$ is a well defined linear functional in $W^*$. Also, it should be easy to verify that $f^*(y^*)=x^*$. Thus, $\ker \Phi={\rm im}(f^*)$. Now, use the first isomorphism theorem to conclude.