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^* &\longrightarrow& V^* \\
f\ &\longmapsto& 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?
 A: 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 $f=g$. 
A: For the case where $T$ is injective, suppose $a \in V^*$. We will define $b \in W^*$ such that $T^*(b) = b\circ T = a$. Let $W = T(V) \oplus W'$, i.e., $W'$ is the complementary subspace of $T(V)$ in $W$. For each $w \in W$, break it into $w = T(v) + w'$ where $v \in V$ and $w' \in W'$. This decomposition is unique (once $W'$ is chosen and is fixed). Since $T$ is injective, the map $w \mapsto v$ is well-defined, and so we can define $b(w) = b(T(v) + w') = a(v)$. It is easy to verify that now $(b \circ T)(v) = b(T(v)) = a(v)$ for all $v \in V$.
For the case where $T$ is surjective, suppose $b \in \ker(T^*)$, i.e., $(b \circ T)(v) = 0$ for all $v$. By surjectivity of $T$, this implies $b(W) = 0$, and so $b = 0$. Therefore, $\ker(T^*) = 0$.
A: 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?
A: For any linear operator $T$, $\ker T = (\mathrm{range}(T^*))^\perp.$
If $T$ is injective, then
$$0 = \ker T = (\mathrm{range}(T^*))^\perp\mathrm{, and} $$ 
$$0^\perp = \mathrm{range}(T^*), \mathrm{so}$$
$T^*$ is surjective. 
The proof that $T$ surjective implies $T^*$ injective is similar. 
