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Given $L$ and $M$ be finite-dimensional vector spaces and let $g:L\to M$ be a linear map. Then there exists a canonical isomorphism from $\operatorname{coker} g^*$ to which of the following spaces - $\ker g $; $(\ker g)^*$.

Progress

The problem is that I read about this in a book, and there the problem asked to construct a canonical isomorphism between $\operatorname{coker} g^*$ and $\ker g$. What I obtained was canonical isomorphism between $\operatorname{coker} g^*$ and $(\ker g)^*$. At most one of these canonical isomophisms could exist, since otherwise we would get a canonical isomorphism between $\ker g$ and $(\ker g)^*$. Please correct me if I am wrong. And tell me whether what I have obtained is correct.

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  • $\begingroup$ Welcome to MSE! What have you tried? What ideas from your recent learning come to mind? $\endgroup$ – Omnomnomnom Sep 20 '14 at 0:46
  • $\begingroup$ Thank you for welcoming and for editing! I've been reading things from here for a long time, and I have a need from help, so I decided to write this time. The problem is that I read about this in a book, and there the problem asked to construct a canonical isomorphism between coker g* and ker g. What I obtained was canonical isomorphism between coker g* and (ker g)*. At most one of these canonical isomophisms could exist, since otherwise we would get a canonical isomorphism between ker g and (ker g)*. Please correct me if I am wrong. And tell me whether what I have obtained is correct. $\endgroup$ – Stoyan Apostolov Sep 20 '14 at 1:03
  • $\begingroup$ Presumably you mean $\ker(g^*)$? I'm not sure what the adjoint of a space would refer to. The book's problem is correct; I'll elaborate in my answer. $\endgroup$ – Omnomnomnom Sep 20 '14 at 1:42
  • $\begingroup$ See my updated answer below. $\endgroup$ – Omnomnomnom Sep 20 '14 at 1:55
  • $\begingroup$ Actually $(\ker g)^*$ stands for the dual space of $\ker g$. Analogously $g^*$ stands for the dual mapping. If we consider these mappings as matrices operating on the same space, then $g^*$ coincides with the transpose of g ($g^T$).However, in order to be accurate, $g^*$ is defined in $M^*$ and taking values in $L^*$. $\endgroup$ – Stoyan Apostolov Sep 20 '14 at 11:33
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$\DeclareMathOperator{\coker}{\operatorname{coker}}$You are correct, the canonical isomorphism is from $\coker g^* \cong (\ker g)^*$, and also that $(\ker g)^*$ and $\ker g$ are not canonically isomorphic.

To see the isomorphism abstractly, consider the exact sequence

$$0 \to \ker g \to L \xrightarrow{g} M \to \coker g \to 0$$

Dualizing (i.e. applying $\text{Hom}_k(\_, k)$ where $k$ is the base field) gives an exact sequence

$$0 \to (\coker g)^* \to M^* \xrightarrow{g^*} L^* \to (\ker g)^* \to 0$$

which precisely identifies $\coker g^*$ with $(\ker g)^*$, and $\ker g^*$ with $(\coker g)^*$. As one can see, this isomorphism is truly natural, and does not require finite-dimensionality.

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