2
$\begingroup$

I am stuck in this exercise.

"Let be $k$ a division ring, and $\textbf{vect-k}$ (resp. $\textbf{k-vect})$ te category of all right (left) vector spaces over $k$ that are of finite dimension, and let D:=$Hom(-, \hspace{0.1cm} _{k}k_{k}):\textbf{vect-k}\rightarrow \textbf{k-vect}$ the usual duality functor (i.e. the controvariant functor that sends every $V_{k}$ in its dual $_{k}V^{*}$ and every linear mapping $f:V_{k}\rightarrow W_{k}$ in its transposed $f^{*}:\hspace{0.1cm }_{k}W^{*}\rightarrow \hspace{0.1cm} _{k}V^{*}$). Show that D is not an isomorphism of categories between $\textbf{vect-k}$ and $(\textbf{k-vect})^{op}$."

$\endgroup$
2
  • $\begingroup$ What are your thoughts on the problem? What have you tried? $\endgroup$ Nov 9 '19 at 14:39
  • $\begingroup$ I was able to show that D is certainly a duality, i.e. a full, faithful and essentially surjective functor, but I've no idea to get out of this question. An isomorphism of categories means that I can find another functor G such that GF and FG are the identities functors (in their respectively domains), so I think I could suppose that such G exist and then find some contraddiction, but I don't see the contraddiction... $\endgroup$
    – Kushike
    Nov 9 '19 at 14:47
0
$\begingroup$

The obstruction lies in cardinality. Prove that: either $V^*$ is finite-dimensional or of uncountable dimension. Therefore, $(\bullet)^*$ cannot be essentially surjective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.