The transpose of a linear injection is surjective.

Solution 1:

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 $f=g$.

Solution 2:

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