Proof

Every Notice that every vector space$W$ is the filtered colimit of the diagram of finite-dimensional subspaces $W' \subseteq W$ and inclusions between them; them. applying Applying this to$W = V$, the condition that $\hom(V,-)$ preserves filtered colimits implies that the canonical comparison map from the followingfiltered colimit over the finite-dimensional subspaces is an isomorphism:

is Therfore an some elementisomorphism$[f]$ , so in some the element colimit represented by$[f<span><del\; class="diffmod">\; ]</del><ins\; class="diffmod">\; :</ins></span>V\to V\prime$ [f] f \colon V \to V' in gets mapped to the colimit represented by$f: V \to V'$identity gets mapped to$1_{\mathrm{id}}$ 1_V id_V, i.e., $i\circ f=1_{\mathrm{id}}$ i \circ f = 1_V id_V for some inclusion $i::V\prime \hookrightarrow V$ i: i \colon V' \hookrightarrow V. This implies $i$ is an isomorphism, so that $V$ is finite-dimensional.

In the converse direction, observe that $\hom(V, -)$ has a right adjoint (and in particular preserves filtered colimits) if$V$hence preserves all colimits) if $V$ is finite-dimensional.

To see this, first notice that the dual vector space $V^\ast$ of functionals $f::V\to k$ f: f \colon V \to k to the ground field is a dual object to $V$ in the sense of monoidal category theory , sense, so that there is acounit ($eva \colon V^\ast \otimes V \to k$evaluation map ) taking$f \otimes v \mapsto f(v)$. The unit is uniquely determined from this counit and can be described using any basis $e_i$ of $V$ and dual basis $f_j$ as the map

taking The$1 \mapsto \sum_i e_i \otimes f_i$unit . We is thus uniquely have determined an from adjunction this counit and can be described using any$(- \otimes_k V) \; \dashv (- \otimes V^\ast)$linear basis , which ismated$e_i$ to of anadjunction$V$ and dual basis$\mathrm{hom}{f}_j(V,-)⊣\mathrm{hom}(V^{*},-)$ \hom(V, f_j -) \dashv \hom(V^\ast, -) by as familiar the hom-tensor map adjunctions; thus$\hom(V, -)$ has a right adjoint.