Let $V$ be a vector space over a field $K$ . Recall that $V^*$ , the dual space, is defined to be the vector space of all linear forms on $V$ . There is a natural embedding of $V$ into $V^{**}$ , the dual of its dual space. In the language of categories, this embedding is a natural transformation between the identity functor and the double dual functor, both endofunctors operating on $\cV_K$ , the category of vector spaces over $K$ .

Turning to the details, let $$\rI, \rD:\cV_K\rightarrow \cV_K$$ denote the identity and the dual functors, respectively. Recall that for a linear mapping $L:U\rightarrow V$ (a morphism in $\cV_K$ ), the dual homomorphism $D[L]:V^*\rightarrow U^*$ is defined by $$D[L](\alpha): u \mapsto \alpha(Lu),\quad u\in U,\; \alpha\in V^*.$$ The double dual embedding is a natural transformation $$\delta:\rI\rightarrow \rD^2,$$ that associates to every $V\in \cV_K$ a linear homomorphism $\delta_V\in\Hom(V,V^{**})$ described by $$\delta_V(v): \alpha\mapsto \alpha(v),\quad v\in V,\;\alpha\in V^*$$ To show that this transformation is natural, let $L:U\rightarrow V$ be a linear mapping. We must show that the following diagram commutes:

Let $u\in U$ and $\alpha\in V^*$ be given. Following the arrows down and right we have that $$(\delta_V\circ L)(u): \alpha\mapsto \alpha(Lu).$$ Following the arrows right, then down we have that \begin{eqnarray*} (\rD[\rD[L]]\circ \delta_U)(u): \alpha &\mapsto& (\delta_U u)(\rD[L]\alpha) \\ &=& (\rD[L]\alpha)(u) \\ &=& \alpha(Lu), \end{eqnarray*}as desired.

Let us also note that for every non-zero $v\in V$ , there exists an $\alpha\in V^*$ such that $\alpha(v)\neq 0$ . Hence $\delta_V(v)\neq 0$ , and hence $\delta_V$ is an embedding, i.e. it is one-to-one. If $V$ is finite dimensional, then $V^*$ has the same dimension as $V$ . Consequently, for finite-dimensional $V$ , the natural embedding $\delta_V$ is, in fact, an isomorphism.