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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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2 Answers 2

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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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I want to post an answer that can be easily understood by physicists. If you find any mistakes please leave a comment.

First of all, suppose $V$ is a $n$-dimensional vector space over the field $\mathbb{k}$, and $W$ is an $m$-dimensional vector space over $\mathbb{k}$. Let $f$ be a $\mathbb{k}$-linear map from $V$ to $W$. i.e. $f\in\mathrm{Hom}_{\mathbb{k}}(V,W)$.

Suppose $p\in\mathrm{Hom}_{\mathbb{k}}(W,\mathbb{k})$, i.e. given an element $p$ of the dual space $W^{\ast}$, then one can define a linear functional on $V$ via the following commutative diagram,

\begin{CD} V @>f>> W\\ @. {_{\rlap{\ p\circ f}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVpV\\ @. \mathbb{k} \end{CD}

and $f$ induces a $\mathbb{k}$-linear map

$$f^{\ast}:W^{\ast}\rightarrow V^{\ast}.$$

To be specific, one can choose a basis $\left\{\mathbf{e}^{V}_{\,1},\cdots,\mathbf{e}^{V}_{\,n}\right\}$ for $V$, and a basis $\left\{\mathbf{e}^{W}_{\,1},\cdots,\mathbf{e}^{W}_{\,m}\right\}$ for $W$. For example, a vector $v\in V$ and a vector $w\in W$ can be written as

$$v=\sum_{i=1}^{n}v^{i}\mathbf{e}^{V}_{\,i}\quad\mathrm{and}\quad w=\sum_{a=1}^{m}w^{a}\mathbf{e}^{W}_{\,a}.$$

Under the linear map $f$, one has

$$f(v)=\sum_{i=1}^{n}v^{i}f(\mathbf{e}^{V}_{\,i})=\sum_{i=1}^{n}\sum_{a=1}^{m}\left(v^{i}F^{a}_{i}\right)\mathbf{e}^{W}_{\,a}.$$

Similarly, one can choose a basis $\left\{\mathbf{e}_{V^{\ast}}^{\,1},\cdots,\mathbf{e}_{V^{\ast}}^{\,n}\right\}$ for $V^{\ast}$, and a basis $\left\{\mathbf{e}_{W^{\ast}}^{\,1},\cdots,\mathbf{e}_{W^{\ast}}^{\,m}\right\}$ for $W^{\ast}$, such that the following identities hold.

$$\mathbf{e}_{V^{\ast}}^{\,i}(\mathbf{e}^{V}_{\,j})=\delta^{\,i}_{\,\,j}\quad\quad\mathbf{e}_{W^{\ast}}^{\,a}(\mathbf{e}^{W}_{\,b})=\delta^{\,a}_{\,\,b}.$$

Then the functional $p\in W^{\ast}$ can be written as

$$p=\sum_{a=1}^{m}p_{a}\mathbf{e}_{W^{\ast}}^{\,a}.$$

Then, one has

$$f^{\ast}(p)(v):=p(f(v))=\sum_{a=1}^{m}p_{a}\mathbf{e}_{W^{\ast}}^{\,a}\left(\sum_{i=1}^{n}\sum_{b=1}^{m}\left(v^{i}F^{b}_{i}\right)\mathbf{e}^{W}_{\,b}\right)=\sum_{a=1}^{m}\sum_{i=1}^{n}\left(v^{i}F_{i}^{a}\right)p_{a}.$$

The transformation rule $p_{a}\mapsto F^{a}_{i}p_{a}$ implies that the maps from vector spaces to dual vector spaces form a contravariant functor.

To show the following diagram

\begin{CD} V @>\phi^{V}>> V^{\ast\ast}\\ @V{f}VV @VV{f^{\ast\ast}}V\\ W @>>\phi^{W}> W^{\ast\ast} \end{CD}

is commutative, first notice that $V^{\ast\ast}=\left(V^{\ast}\right)^{\ast}$, and $W^{\ast\ast}=\left(W^{\ast}\right)^{\ast}$.

This means that for any $v\in V$, under the map $\phi^{V}:V\rightarrow V^{\ast\ast}$, one has $\phi^{V}(v)\in\mathrm{Hom}_{\mathbb{k}}(V^{\ast},\mathbb{k})$. i.e. for any $q\in V^{\ast}$, one has $\phi^{V}(v)(q)\in\mathbb{k}$. It can be naturally defined via

$$\phi^{V}(v)(q):=q(v),$$

for any $v\in V$ and $q\in V^{\ast}$. Similarly, one can define $\phi^{W}$ via

$$\phi^{W}(w)(p):=p(w),$$

for any $w\in W$ and $p\in W^{\ast}$.

But as mentioned earlier, covectors in $V^{\ast}$ are defined via the induced map $f^{\ast}:W^{\ast}\rightarrow V^{\ast}$. So a generic covector $q\in V^{\ast}$ should take the form $q=f^{\ast}(p)=p(f)$, for some $p\in W^{\ast}$. So it is natural to consider the map

$$\phi^{V}(v)\circ f^{\ast}:W^{\ast}\rightarrow\mathbb{k}.$$

One can notice that $\phi^{V}(v)\circ f^{\ast}\in W^{\ast\ast}$, and so the map $f^{\ast}:W^{\ast}\rightarrow V^{\ast}$ induces a covariant map $f^{\ast\ast}: V^{\ast\ast}\rightarrow W^{\ast\ast}$ defined as follows:

$$f^{\ast\ast}\circ\phi^{V}(v)=f^{\ast\ast}(\phi^{V}(v)):=\phi^{V}(v)\circ f^{\ast}.$$

Specifically, for a given vector $v\in V$ and a covector $p\in W^{\ast}$, one has

$$f^{\ast\ast}\circ\phi^{V}(v)(p)=\phi^{V}(v)\circ f^{\ast}(p)=\phi^{V}(v)(f^{\ast}(p))=f^{\ast}(p)(v)=p(f(v)).$$

Similarly, using the map $f:V\rightarrow W$, one has

$$\phi^{W}(f(v))(p)=p(f(v)).$$

Therefore, one concludes that

$$f^{\ast\ast}\circ\phi^{V}=\phi^{W}\circ f.$$

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