# Inclusion functor $\mathbf{FinVect}_K \to \mathbf{Vect}_K$ has no adjoint

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 $R$?

• 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. – Derek Elkins Jun 8 '17 at 21:24
• @DerekElkins I'm not sure I see the relevance of infinite coproducts to a question about finite dimensional vector spaces. – Kevin Carlson Jun 9 '17 at 7:21
• @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. – 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.