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$\require{AMScd}$ Note: I have already seen this question, which asks about a specific aspect of the construction - here I am trying to construct this functor and failing at a very different stage.

We are trying to show that the two functors $\sf Id:FDVec_k\to FDVec_k$ and $(\cdot)^{\vee\vee}:\sf FDVec_k\to FDVec_k$ are naturally isomorphic. I've never seen such an argument before, and working through the linear algebra is giving me some problems.

An element of $V^{\vee\vee}$ is defined to be a linear map from $V^{\vee}$ to $k$, so it takes linear maps $V\to k$ to elements of $k$. For a natural isomorphism we need isomorphisms $$m_V:V\to V^{\vee\vee}.$$ We input a vector and output a map $\operatorname{Hom}(V,k)\to k$: $$m_V(v)(f:V\to k)=f(v)\in k.$$ We then need to show that diagrams $$ \begin{CD} V @>\phi>> W\\ @V{m_V}VV @VV{m_W}V\\ V^{\vee\vee} @>>\phi^{\vee\vee}> W^{\vee\vee} \end{CD} $$

commute if $\phi$ is a linear map of finite-dimensional vector spaces, so $$m_W\circ\phi=\phi^{\vee\vee}\circ m_V.$$

However, my problem ultimately ends up being that I can't interpret $\phi^{\vee\vee}$ sufficiently well. It should take linear maps $\operatorname{Hom}(V,k)\to k$ to linear maps $\operatorname{Hom}(W,k)\to k$ in a way that is somehow induced by $\phi$, but I'm not sure what this way is. My attempt:

$$[\phi^{\vee\vee}(f:V^{\vee}\to k)](g:W\to k)\in k$$

but I get no further than this. I ask purely for clarification as to what a double dual map really is (the several layers of abstraction present in the problem are quite hard for my beginner's mind to handle).

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You only need to understand what the first dual map $\phi^\vee$ is, as $(\phi^\vee)^\vee=\phi^{\vee\vee}$. The map $\phi^\vee$ is defined by $\phi^\vee(f)=f\circ\phi$. This is just a special case of the contravariant hom functor. To check that the diagram is commutative, notice that for $v\in V$ and $f\in W^\vee$ we have $$(\phi^{\vee\vee}\circ m_V)(v)(f)=(\phi^{\vee\vee}(m_V(v))(f)=(m_V(v)\circ\phi^\vee)(f) \\ =m_V(v)(f\circ\phi)=f(\phi(v))=m_W(\phi(v))(f).$$

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