# Why is the the double dual functor on finite-dimensional vector spaces naturally isomorphic to the identity?

$\require{AMScd}$ 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).

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 $$(\phi^{\vee\vee}\circ m_V)(v)(f)=(\phi^{\vee\vee}(m_V(v))(f)=(m_V(v)\circ\phi^\vee)(f) \\ =m_V(v)(f\circ\phi)=f(\phi(v))=m_W(\phi(v))(f).$$

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

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