Isomorphisms between a vector space and its dual For finite dimensional vector spaces $V$ and $W$, let $i_V: V \rightarrow V^{**}$ and $i_W: W \rightarrow W^{**}$ be natural isomorphisms. Show that for any linear transformation $f : V \rightarrow W$, $i_W\circ f = f^{**}\circ i_V$. And in addition, show that there do not exist isomorphisms $i_V : V \rightarrow V^*$ and $i_W : W \rightarrow W^*$ such that for any linear transformation $f: V\rightarrow W$, such that $f^{*}\circ i_W\circ f = i_V$.
Questions: For the first part of the question, are $f$ and $f^{**}$ related in any way? Similarly, for the second part of the question, are $f$ and $f^*$ related? In addition, can I assume a basis for $V$ and a basis for $W$ to solve these questions? Any hints? For the second part, where to start?
Update:


*

*Part $1$: For an arbitrary element $v\in V$, the left hand side is $i_W[f(v)\in W]\in W^{**}$. That is, for an arbitrary $\psi\in W^*$, $i_W[f(v)\in W](\psi) =\psi[f(v)] \in\mathbb R$. Now consider the right hand side, for an arbitrary $\phi\in V^*$, $i_V(v)(\phi)=\phi(v)\in V^{**}$; then $f^{**}[i_V(v)]\in W^{**}.$ Hence, $f^{**}[i_V(v)](\psi)= i_V(v)[f^*(\psi)\in V^*]=f^*(\psi)(v)\in\mathbb R.$ I should now have $f^*(\psi)(v) = \psi[f(v)] $ simply by definition of $f^*$.

*Part $2$: Assume there exist the two isomorphism $i_V$ and $i_W$. Let $V=W=\mathbb R^2$ and set $f$ to be $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.$ Consider the left hand side, for an arbitrary $(a, b)\in \mathbb R^2$, $f(a, b) = (b, a)$. Then $i_W(b, a)\in W^*$ and $f^*[i_W(b, a)](a, b) = i_W(b, a)[f(a,b)] = i_W(b, a)(b, a)\in\mathbb R$. As for the right hand side, $i_V(a, b)(a,b)\in\mathbb R$. I should now have $i_V(a, b)(a,b)\neq i_W(b, a)(b, a)$, but how?
 A: The maps $f^\star$ and $f^{\star\star}$ are the dual and double-dual of $f$, respectively. So, "Yes, they're related to $f$". 
In particular, for $\phi$ in $W^\star$, the definition of $f^\star$ is something like this:
$f^\star(\phi)$ is supposed to be an element of $V^\star$, so it takes elements of $V$ to real numbers. So for $v \in V$, what's $f^\star(\phi)(v)$? it's just
$$
f^\star(\phi)(v) = \phi(f(v)).
$$
The double dual is this applied twice. 
You could assume bases for $V$ and $W$, but it's not needed, as Hagen suggests. 
For the second part, I believe that one of the two following approaches will get you somewhere: 


*

*Pick $V = W = \mathbb R^2$. Go ahead and pick bases, and suppose that you've got the isomorphisms. Now consider maps $f$ of the form
$$
\begin{bmatrix}
1 & a \\
0 & 1
\end{bmatrix} \\ \text { and } \\
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}.
$$

*Pick $ V = \mathbb R$ and $W = \mathbb R^2$ (or vice versa) and consider either the maps 
$1 \mapsto a e_1 + b e_2$ or the maps $a e_1 + b e_2 \mapsto ap + bq$, accordingly. I think that one of these will lead you to a contradiction, but I'm not certain. 
Addendum: solution to part 2:
$$
\newcommand{\vstar}{V^{\star}}
\newcommand{\wstar}{W^{\star}}
\newcommand{\fstar}{f^{\star}}
\newcommand{\iv}{i_V}
\newcommand{\iw}{i_W}
\newcommand{\Rtwo}{{\mathbb R^2}}
\newcommand{\aij}{a_{ij}}
\newcommand{\bij}{b_{ij}}
\newcommand{\cij}{c_{ij}}
$$
$$
\newcommand{\Amat}{\mathbf A}
\newcommand{\Bmat}{\mathbf B}
\newcommand{\Cmat}{\mathbf C}
\newcommand{\vvec}{\mathbf v}
\newcommand{\evec}{\mathbf e}
\newcommand{\ei}{\mathbf e_i}
\newcommand{\ej}{\mathbf e_j}
\newcommand{\ek}{\mathbf e_k}
\newcommand{\el}{\mathbf e_\ell}
$$
For part 2, let’s work with $V = W = \Rtwo$ and the standard basis $\evec_1, \evec_2$ for $\Rtwo$. All indices in what follows will be over the set $\{1, 2\}$, so when I write “Let $\phi_i = \iv(\ei)$”, I mean  “Let $\phi_1 = \iv(\evec_1)$ and $\phi_2 = \iv(\evec_2)$.” Clear? 
I need one small observation before I start: for any nonzero vector $\vvec \in V = \Rtwo$, there’s a functional $\phi$ with $\phi(\vvec) \ne 0$. (For instance, using dot-products, we could define $\phi(\mathbf x) = \vvec \cdot \mathbf x$.)
The proof is by contradiction, i.e., I’ll assume the existence of $i_V$ and $i_W$ and derive a contradiction. 
OK. Let
$$
\phi_i = \iv(\ei) \\
\psi_i = \iw(\ei).
$$
These constitute bases $\Phi$ and $\Psi$ for $\vstar$ and $\wstar$, respectively. Why are they bases? Because each is the image, under an isomorphism, of a basis for $\Rtwo$.  
Let
$$
a_{ij} = \phi_i (\ej).
$$
Then the numbers $a_{ij}$ are the entries of a matrix $\Amat$, which is nonsingular. Proof: suppose that $\Amat\vvec = 0$, but $\vvec \ne 0$.  Then for each $i$, 
$$
\sum_j a_{ij}v_j = 0 \\
\sum_j \phi_i(\ej)v_j = 0 \\
\phi_i(\sum_j v_j\ej) = 0 \\
\phi_i(\vvec) = 0 
$$
In other words, both $\phi_1$ and $\phi_2$ kill $\vvec$. Since the $\phi$s form a basis, every element of $\vstar$ would kill $\vvec$. This contradicts the observation above.
Similarly, the numbers $\bij = \psi_i(\ej)$ form a nonsingular matrix $\Bmat$. 
Finally, we can write $f(\ei) = \sum_j \cij \ej$. The matrix $C$ for the transformation $f$ is not necessarily nonsingular, however. 
With these conventions, we can say something about the matrices of $\iv, \iw, f, $ and $\fstar$ with respect to the various bases. I’ll write $T|_{K,L}$ for the matrix of the transformation $T$ with repect to the bases $K$ and $L$ of its domain and codomain, respectively. I’ll use $E$ to denote the basis $\{\evec_1, \evec_2 \}$ of $V = W = \Rtwo$. 
The definitions of $\phi$ and $\psi$ show that 
$$
\iv|_{E, \Phi} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\
\iw|_{E, \Psi} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\
$$
and the matrix of $f$ is 
$$
f|_{E, E} = \Cmat.
$$
Now, let’s study $\fstar( \iw(f(\ei) )$, i.e., the image of $\ei$ under the left, bottom, and right maps in the diagram
$$
\begin{matrix}
V & \stackrel{i_V}{\longrightarrow}&  V^{\star}\\
f\downarrow & &  \uparrow f^{\star} \\
W & \stackrel{i_W}{\longrightarrow}&  W^{\star}\\
\end{matrix}
$$
that we're supposing exists and is commutative. 
Now $\fstar( \iw(f(\ei) )$ is an element of $\vstar$, so we can study it by looking at its value on an arbitrary vector $\vvec \in V$, say $\vvec = \sum_k v_k \evec_k$. We get
\begin{align}
\fstar( \iw(f(\ei) )) (\vvec) &= \iw (f(\ei) ) (f(\vvec)) \text{, by defintion of the dual $\fstar$}\\
&= \iw (\sum_j \cij \ej ) (f(\sum_k v_k \ek)) \\
&= \iw (\sum_j \cij \ej ) (\sum_k v_k f(\ek)) \text{, by linearity of $f$}\\
&= \iw (\sum_j \cij \ej ) (\sum_k v_k \sum_\ell c_{k \ell} \el) \text{, by formula for $f$}\\
&= (\sum_j \cij \psi_j ) (\sum_k v_k \sum_\ell c_{k \ell} \el) \text{, by definition of $\psi_i$}\\
&= \sum_{jk\ell} \cij v_k c_{k\ell} \psi_j (\el) \\
&= \sum_{jk\ell} \cij v_k c_{k\ell} b_{j \ell} \text{, definition of $\bij$}\\
&= \sum_{jk\ell} \cij b_{j \ell}c_{k\ell}v_k  \text{, rearrangement of factors} \\
&= ((\Cmat\Bmat\Cmat^{t})\vvec)_i
\end{align}
so that in summary, we get the $i$th entry of $(\Cmat\Bmat\Cmat^{t})\vvec$.
On the other hand, reading across the top of the diagram, this must be equal to 
\begin{align}
\iv((\ei) ) (\vvec) &= \phi_i (\sum_k v_k \ek) \\
&= \sum_k a_{ik} v_k\\
&= (\Amat \vvec)_i
\end{align}
Because this is true for $i = 1, 2$, and for any vector $\vvec$, we must have 
$$
\Amat = \Cmat\Bmat\Cmat^{t}
$$
But $\Cmat$ is the matrix for the (arbitrary) transformation $f$. Picking $f = $identity, we get that $\Amat = \Bmat$, but picking $f = 2\mathbf I$, we get $\Amat = 4 \Bmat$, which can only happen if $\Amat = \Bmat = 0$, which is a contradiction, because both are nonsingular. 
I apologize for how convoluted this all is. I’m certain that there’s a simpler proof lurking somewhere in there. 
A: Here's an improved answer to part 2, at least for the case where $V$ and $W$ are finite dimensional vector spaces of different dimensions. Let's suppose $dim V = n$ and $dim W = k$. You're hoping to have $f^{\star} \circ i_W \circ f = i_V$ for every $f$. Well, if $k < n$, then the rank of $f$ is at most $k$, so the rank of the composite map on the left is at most $k$, while the rank of $i_v$ is $n$. That shows that the existence of $i_V$ and $i_W$ in general is impossible. 
Assuming that $i_W$ is supposed to depend only on $W$, and not on $V$, we can always pick $V = W \times  \mathbb R$, so that the dimension of $V$ is one greater than the dimension of $W$, so that $i_W$ cannot exist. 
A: The first part is to show that $ V\rightarrow V^{**}$ is an natural transformation and the second to show that $ V\rightarrow V^{*}$ is not a natural transformation.
First Part: Define $i_V : V\rightarrow V^{**}$ by 
$$i_V(x)(\alpha)=\alpha(x)$$
Notation: $x,y\ldots$ denote vectors in $V, W$ and $\alpha, \beta\ldots$ denote linear functionals in  $V^{*}, W^{*}$. So $i_V(x)$ is a functional on $V^{*}$ and its value at $\alpha$ is $\alpha(x)$. 
Now let $f:V\rightarrow W$ be a linear map then define $f^{*}:W^{*}\rightarrow V^{*}$
by 
$$f^{*}(\alpha)(x)=\alpha(f(x))$$ where $\alpha \in W^{*}$
and $x\in V$.
The equation to be shown is 
$$i_W\circ f=f^{**}\circ i_V$$
let $x \in V$ then we need the equality
$$i_W(f(x))=f^{**}(i_V(x))$$ in $W^{**}$.
So let $\alpha \in W^{*}$ then we need the equality in the base field
$$i_W(f(x))(\alpha)=f^{**}(i_V(x))(\alpha)$$
Now for the left side
$$i_W(f(x))(\alpha)=\alpha(f(x))$$ is immediate and for the other side,
\begin{equation*}
\begin{split}
f^{**}(i_V(x))(\alpha)&=i_V(x)(f^{*}(\alpha))\\
&=f^{*}(\alpha)(x)\\
&=\alpha(f(x))\\
\end{split}
\end{equation*}
Second Part: Let $i_V : V\rightarrow V^{*}$ and $i_W : V\rightarrow W^{*}$ be given isomorphisms. Then they define non-degenerate bilinear forms on $V$ and $W$
via $$(y,x)=i_V(x)(y)$$ for $x,y \in V$ and similarly for $W$.
Conversely an non-degenerate bilinear forms defines an isomorphism with the dual.
Now the equation of interest is 
$$f^{*}\circ i_W \circ f=i_V.$$
For $x\in V$ this means 
$$f^{*}(i_W (f(x)))=i_V(x)$$ holds in $V^*$ which in turn means that for any $y \in V$,
$$f^{*}(i_W (f(x)))(y)=i_V(x)(y)$$ 
or unraveling the definitions again,
$$i_W (f(x))(f(y))=i_V(x)(y).$$
which in terms of inner products means that 
$$(y,x)=(f(y),f(x)).$$
In other words the desired equation is equivalent to saying that $f$ is an orthogonal transformation.
 Now under all these translations the naturality statement reads:
If $V$ and $W$ are two non-degenerate bilinear spaces and $f:V \rightarrow W$ is a linear map then $f$ is orthogonal.
However we see this now as false, it is easy to construct a counterexample in all but the dimension one case.
We need a map $f:V \rightarrow W$ which is not othogonal.
Let us note the following fact relative to any non degenerate bilinear space.
If there is such that $(x,x) \neq 0$ then there is $y$, independent from $x$,
with $(x,y)=0$. This follows from the following equation 
$$(x,y-\frac{(x,y)}{(x,x)}x)=(x,y)-\frac{(x,y)}{(x,x)}(x,x)=0.$$
Now to find $f$ first assume that there is $x \in V$ such that $(x,x)\neq 0$, then we have independent $x,y \in V$ with $(x,y)=0$ now there are $u,w \in W$ such that 
$(u,w)\neq 0$ by the assumption of non degeneracy. So $f(x)=u$ and $f(y)=w$ is an example of a non orthogonal map.
Now assume that $(x,x)=0$ for all $x \in V$. We have linearly independent $x,y$ such that $(x,y)\neq 0$ (because some such pair exists and they cannot be dependent by the assumption of this case). Now note that there are $u,w 
\in W$ such that $(u,w)=0$, since either $(u,u)=0$ all $u$ or by the argument above there are two orthogonal vectors. So again $f(x)=u$ and $f(y)=w$ is an example of a non orthogonal map.
