An example of a linear transformation that is not 1-1 or onto but has bijective dual This is a question from some ring theory problems in Linear Algebra.
I am not sure if this question is simple enough to come up with a linear map on a finite dimensional vector space but here it goes. 

Give an example of a linear map between two vectors spaces such that the dual is bijective but the original map is neither 1-1 or onto.

If the vector spaces in question have finite dimension we would require the matrix representing the transpose to have determinant zero but the transpose to be singular so that is why I am thinking there must be an infinite dimensional example.
 A: I'll follow Wikipedia's terminology: If $f:V\to W$ is a linear map, then the transpose (or dual) $f^*:W^*\to V^*$ is defined by $$f^*(\varphi)=\varphi\circ f.$$ 
The first question is: Give an example of a linear map between two vectors spaces such that the transpose is bijective but the original map is neither 1-1 nor onto. 
There are no such examples. Indeed, in the above notation, if $f:V\to W$ is not injective, then $f^*:W^*\to V^*$ is not surjective, because the image of $f^*$ consists of those linear forms on $V$ which vanish on $\ker f$. Similarly, if $f$ is not surjective, then $f^*$ is not injective, because the kernel of $f^*$ consists of those linear forms on $W$ which vanish on $f(V)$. 
Another question was added in the comments: Are there still no examples if, instead of vector spaces, we consider modules over a commutative ring $R$ with identity?
Such examples exist. A cheap one is given by setting $R=\mathbb Z$, $V=W=\mathbb Z/2\mathbb Z$, $f=0$. 
EDIT. In fact we have canonical isomorphisms
(1) $\text{Ker}(f^*)=\text{Coker}(f)^*$ and $\text{Coker}(f^*)=\text{Ker}(f)^*$. 
This can bee seen as follows: 
(2) If
$$
0\to A\overset{i}{\to}V\overset{f}{\to}W\overset{p}{\to}B\to0
$$ 
and 
$$
0\to A'\overset{i'}{\to}V\overset{f}{\to}W\overset{p'}{\to}B'\to0
$$ 
are exact sequences of $K$-vector spaces, then there is a unique linear map $a:A\to A'$, and a unique linear map $b:B\to B'$ such that $i'\circ a=i$ and $b\circ p=p'$. Moreover $a$ and $b$ are bijective. The proof is easy. 
(3) We have the exact sequences 
$$\begin{matrix}
0\to&\text{Ker}(f)&\overset{i}{\to}V\overset{f}{\to}W\overset{p}{\to}&\text{Coker}(f)&\to0,\\ \\ 
0\to&\text{Coker}(f)^*&\overset{p^*}{\to}W^*\overset{f^*}{\to}V^*\overset{i^*}{\to}&\text{Ker}(f)^*&\to0,\\ \\ 
0\to&\text{Ker}(f^*)&\to W^*\overset{f^*}{\to}V^*\to&\text{Coker}(f^*)&\to0.
\end{matrix}
$$ 
Then (1) follows from (2) and (3). 
