Prove $\dim {\rm Im}\ T = \dim W$ in the proof of the dimension theorem for the symplectic complement

Question

I am trying to prove the dimension theorem for the symplectic complement and I am missing a small step right at the end:

Let $ (V,\omega) $ be a symplectic vector space and $W$ a linear subspace of $V$. The $\omega$-orthogonal complement is defined to be the linear subspace:

$$W^\omega:=\{v\in V| \omega(v,w)=0, \forall w \in W\}\subseteq V$$ then $$\dim W +\dim W^\omega = \dim V$$

Hint: use $i_\omega$ (defined in the proof) and $j^* $ ($j$ is the canonical inclusion)

Proof: Following the hint:

Considering the map $i_\omega:V\to V^*, i_\omega(v)=\omega(v,\cdot)$ which is shown to be bijective, because $\omega$ is nondegenerate by definition of symplectic form and considering the image of $W^\omega$ under $i_\omega$ and the dual of the canonical inclusion $j^*:V^*\to W^*$ I have the following map:

$$T:= j^*\circ i_\omega :V\to V^*\to W^*: v \mapsto i_\omega(v)\mapsto j^*(i_\omega(v))$$

I have already proven that $\ker T=W^\omega$, so $\dim\ker T = \dim W^\omega$

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Proof of $\ker T=W^\omega$: Let $v \in \ker T \iff Tv=0 \iff 0=Tv= j^*\circ i_\omega(v) = i_\omega(v)\circ j \iff ( i_\omega(v)\circ j)(w) = \omega(v,j(w))=\omega(v,w),\; \forall w \in W,$

This is precisely the definition of $W^\omega$ $\iff v \in W^\omega$

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so if I can prove that $\dim {\rm Im}\ T = \dim W$, I'd get the thesis by the rank-nullity theorem

since ${\rm Im}\ T\subseteq W^*$, $\dim {\rm Im}\ T \leq \dim W^*= \dim W$

How do I prove the opposite inequality?

I am not very sure if I can use this argument: the inclusion map is an injective linear map so the composition $T$ is an injective linear map(but this seems to be in contradiction with $\ker T = W^\omega$ instead of $\{0\}$, so $T$ is not injective?), since a linear map is injective iff it is surjective, then it must be surjective, then $\dim {\rm Im} \ T=\dim W^*$??

How do I solve this contradiction?

EDIT

I find quite strange that $\ker T=W^\omega$ and the $\ker i_\omega={0}$ are different things, after all the $\ker T$ are the vectors of $V$ mapped to the $0 $vector of $W^*$ , the $\ker i_\omega$ are the vectors of $V$ mapped to the $0$ vector of $V^*$ and the $0$ vector of $V^*$ and the zero vector of $W^*$ should be the same, since $V^*$ is a subspace of $W^*$. What is going on?

There's a similar proof here https://planetmath.org/dimensiontheoremforsymplecticcomplementproof but I don't understand the last 2 sentences.

Answer 1

A linear map is injective if and only if its dual is surjective. Thus, $j^*$ is surjective and then so is $T$, being a composition of surjective maps. Hence $\operatorname{im} T = W^*$ and $$\dim \operatorname{im} T = \dim W.$$

Regarding your question in the comments: There is no contradiction, since the equivalence “injective if and only if surjective” only holds for linear maps whose domain and codomain have the same dimension.