Let $f:V\rightarrow W$ be a surjective linear map with $U=\ker(f)\subset V$. Let $j_U:U\hookrightarrow V$ be the inclusion of $U$ in $V$.
How do I show that: $\text{im} f^\top=\ker j_U^\top$?
I am allowed to use the fact that if $f$ is injective, then $f^\top$ is surjective and if $f$ is surjective, then $f^\top$ is injective. But I am not allowed to use annihilators.
For easier typing, let us use ${}^*$ instead of ${}^\top$ and $j$ instead of $j_U$.
The inclusion $\text{im} \ f^* \subseteq \ker j^*$ is trivial: if $\omega \in \text{im} \ f^*$, there exist $\eta \in W^*$ such that $\omega = f^* (\eta)$, so $j^* (\omega) = j^* (f^* (\eta)) = (f \circ j)^* (\eta) = 0^* (\eta) = 0$, so $\omega \in \ker j^*$ (note that $\text{im} \ j = \ker f$ implies $f \circ j = 0$).
The inclusion $\ker j^* \subseteq \text{im} \ f^*$ is slightly subtler, its proof is essentially the proof of the fundamental isomorphism theorem for modules: let $\omega \in \ker j^*$; we shall construct a morphism $\eta \in W^*$ such that $\omega = f^* (\eta)$. Let $w \in W$; since $f$ is surjective, there exist $v \in V$ (not necessarily unique!) such that $f(v) = w$. Define $\eta (w) = \omega (v)$; let us show that $\eta$ is well defined, i.e. let us show that if $v' \in V$ is another element such that $f(v') = w$, then $\omega (v') = \omega (v)$. Note that $f(v' - v) = f(v') - f(v) = w - w = 0$, so $v' - v \in \ker f = \text{im} \ j$, so there exist $u \in U$ such that $v' - v = j(u)$, i.e. $v' = v + j(u)$, so $\omega (v') = \omega (v) + (\omega \circ j) (u) = \omega (v) + j^* (\omega) (u) = \omega (v)$ because $\omega \in \ker j^*$ by assumption. Thus, $\eta$ is well defined. It is fairly easy to show that $\eta$ is a morphism (if you have troubles with this, tell me in a comment). Finally, for $v \in V$, $f^*(\eta) (v) = \eta ( f(v) ) = \omega (v)$ (by the definition of $\eta$), which means that $\omega = f^* (\eta)$.