The transpose of a linear injection is surjective. Let 
$$T:V\longrightarrow W$$
be a linear map (of vector spaces), and let
\begin{eqnarray}
T^*:W^* &amp;\longrightarrow&amp; V^* \\
f\ &amp;\longmapsto&amp; f\circ T
\end{eqnarray}
be its transpose (or dual map). 

How to show that if $T$ is injective then $T^*$ must be surjective, and that if $T$ is surjective then $T^*$ must be injective?
If $T$ is injective then it has a inverse $T^{-1}:T(V)\rightarrow V$. Take $v\in V^\star$ and define $f\in W^\star$ by $f(x)=v(T^{-1}(x))$. Then $T^\star(f)=f\circ T=v\circ T^{-1}\circ T=v$. Hence,$T^{*}$ is surjective.

On the other hand, suppose that  $T$ is surjective. If $T^\star f=T^\star g$, then $f(T(x))=g(T(x))$ for all $x\in V$. Hence $(f-g)(T(x))=0$ for all $x\in V$. Because $T$ is surjective you can conclude that $f-g=0$, or $f=g$.
Partial Solution:

$T$ is injective $\implies T^*$ is surjective: Choose $f\in V^*.$ We need to find out (using the injectivity of $T$) a linear transformation $f':W\to F$ such that $f'T=f.$  Let's try to reduce the proposed problem a bit more using the fact that two linear transformations agree everywhere if and only if they agree on a basis of the  vector space considered as the domain. Thus our reduced task is to find out a linear transformation $f':W\to F$ such that $Tf'=f$ on a chosen basis of $V.$  The problem can a little more be reduced to define $f'$ using the fact that any mapping defined on a basis of a vector space extends to a unique linear transformation. Thus our aim is to define $f'$ on a basis of $W$ such that the effect of the composed map $f'T$ on the basis is identical with that of $f.$ How can it be done? Naturally by considering the effect of the pre-image when it exists (and here the injectivity comes into play) and nullify the effect otherwise. Can you construct $f'$ now?