Let $K$ be a field, $\mathbf{FinVect}_K$ be the category of finite-dimensional $K$-vector spaces and $\mathbf{Vect}_K$ be the category of all $K$-vector spaces.

Why does the natural inclusion functor $\mathbf{FinVect}_K \to \mathbf{Vect}_K$ not have a left or right adjoint? Note that this functor is exact, so we can't use the theorem that a left/right adjoint functor is right/left exact.

What about the more general case, when we consider the inclusion functor from finitely generated left $R$-modules to the category to $_{R}\mathbf{Mod}$ for a non-zero ring $R$?

  • $\begingroup$ If a functor preserves finite (co)limits, but not all (co)limits, then the only remaining case to check to see if the functor is (co)continuous is infinite (co)products. $\endgroup$ – Derek Elkins Jun 8 '17 at 21:24
  • 1
    $\begingroup$ @DerekElkins I'm not sure I see the relevance of infinite coproducts to a question about finite dimensional vector spaces. $\endgroup$ – Kevin Carlson Jun 9 '17 at 7:21
  • 2
    $\begingroup$ @KevinCarlson A left adjoint to this inclusion would be cocontinuous and thus would need to preserve the infinite coproducts in $\mathbf{Vect}_K$. This is the basis of the accepted answer. $\endgroup$ – Derek Elkins Jun 9 '17 at 15:26

If there were an adjoint to this functor, the following would be true: for any vector space $V$, there is a finite-dimensional vector space $W$ and a linear transformation $S: V \to W$ such that if $T: V \to U$ is a linear transformation into a finite-dimensional vector space $U$, then $T$ factors uniquely through $W$ via $S$. That is, $T = T' \circ S$ for a unique $T': W \to U$.

Now take $V$ to be infinite-dimensional and $U = K$. Then any linear functional $V \to K$ induces a unique linear functional $W \to K$. Furthermore, this inducement is easily checked to be $K$-linear (use uniqueness). So we obtain $\text{Hom}_K(V,K) \cong \text{Hom}_K(W,K)$, which cannot be true by considering dimension.

  • 2
    $\begingroup$ To be clear, here you are specifically considering a left adjoint. For a right adjoint the maps would be going the other direction (but the same argument would work). $\endgroup$ – Eric Wofsey Jun 9 '17 at 4:24
  • $\begingroup$ Yes, a good observation. $\endgroup$ – Mr. Chip Jun 9 '17 at 13:04

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.