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?

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, $\ker(T^*) = 0$.
