A space whose dual is $F^m$ Is there a space whose dual is $F^m$? ($F$ is the field w.r.t. which the original set is a vector space)
I'm trying to do the following exercise:

I only need help with the direction from $\Gamma$ is injective.
I was thinking of using the theorem that states that if a dual map is injective then the linear map must be surjective. And if a linear map $T$ from $\mathcal{L}(F^m,V)$ with $T(e_i)=v_i$ is surjective, then the $v_i$ span $V$.
Let's assume that $\Gamma^*(\phi)=(\phi(v_1),...,\phi(v_m))$, where $v_i \in V$, and $\phi \in \mathcal{L}(V,F)$. Is there a way to think of $\Gamma^*$ as the dual map of a linear map $\Gamma$?
 A: Let $e_1, \dots, e_m$ a basis of $F^m$. 
Let $f:F^m \to F$ be a linear form, that is an element of the dual space. 
Then for any $v= a_1e_1 + \dots + a_me_m$ we have $f(v)= f(a_1e_1 +\dots + a_me_m) = a_1f(e_1) +\dots + a_mf(e_m)$. Thus $f$ is uniquely determined by 
$f(e_1), \dots, f(e_m)$ that is an $m$  elements from $F$, viz. an element of $F^m$. 
Conversely for every $(b_1, \dots, b_m) \in F^m$ we can define an associated linear form via defining $g(e_i)=b_i$ and thus $g(a_1e_1 +\dots + a_me_m) = a_1b_1 +\dots + a_mb_m$.
It follows the elements of the dual space are in bijection with the elements of $F^m$. 
It is not hard to show that this bijection is linear, and thus an isomorphism.
Note: the bijection depends on the basis. This is why one says there is no canonical isomorphism, but there still is an isomorphism.  
However the spaces are not literally equal. The one is made up of  $m$-tuples of elements of $F$ the other of linear maps from $F^m$ to $F$, that is certain subsets of the Cartesian product of $F^m \times F$. 
However, when somebody says "the dual of $V$ is {something}", then usually this {something} is just isomorphic to the dual not literally the sets of linear maps from $V$ to $F$. 

For the supplementary question that motivated the question: 
Suppose the function $\Gamma$ is injective. This means that the only form that vanishes on all of $v_1, \dots , v_m$ is the form that maps the whole space to $0$. But this means that $v_1, \dots, v_m$ must span $V$, for if not then we could define a form by saying it is $0$ on the span of $v_1, \dots, v_m$ and $1$ for $w_1, \dots, w_n$ where $w_1, \dots, w_n$ is an independent set that completes $v_1, \dots, v_m$  to a generating set of the space. 
A: Given a vector space $\mathbf F^m$, this space has its double dual space ${({\mathbf F^m})^{\ast\ast}} = \mathcal L(({\mathbf F^m})^{\ast},\mathbf F),$ the set of linear functionals on $({\mathbf F^m})^{\ast}$. This double dual is naturally isomorphic to $\mathbf F^m$. Let $v \in \mathbf F^m$, $\varphi \in ({\mathbf F^m})^{\ast}$ and define $\operatorname{ev}_v:({\mathbf F^m})^{\ast}\to\mathbf F$, the evaluation at $v$ functional on $({\mathbf F^m})^{\ast}$ by
$$
\operatorname{ev}_v(\varphi) = \varphi(v) \in \mathbf F^m.
$$
Note that $\operatorname{ev}_v$ is an element of the double dual of $\mathbf F^m$. The association of $\operatorname{ev}_v$ with $v$ is so natural, they can be thought of as the same element and $\operatorname{ev}_v\mapsto v$ is an isomorphism ${({\mathbf F^m})^{\ast\ast}}\cong \mathbf F^m$. In this sense, the dual space of $\mathbf F^m$ is $({\mathbf F^m})^{\ast}$ since the association $\operatorname{ev}_v\mapsto v$ is so natural. This isomorphism $\operatorname{ev}_v\mapsto v$ is natural in the sense that it does not depend on our choice of basis for $\mathbf F^m$.
However, a vector space $V$ can never truly be equal to its dual for the reason that $V^\ast$ is the set of linear functionals on $V$ and $V$ is itself a set of vectors. This association of $V$ with its double dual via the evaluation map is the closest thing to a vector space equalling its dual.
