The problem arises when I was reading the book Hopf Algebra written by Eiichi Abe. I asked a qustion related to this question on this site Topology of vector spaces.
In the book, a subspace is referred to a topological subspace and a $k$-linear subspace is just a linear subspace.
On page 70, there is an exercise :
Let $u:V\rightarrow W$ be a $k$-linear map and let $u^*:W^*\rightarrow V^*$ be its dual $k$-linear map. Then $u^*$ is a continuous function and the image of a closed subspace of $W^*$ is a closed subspace of $V^*$. Moreover, if we let $S$, $T$ be $k$- linear subspaces of $V$, $W$ respectively, then $$u^{*-1}(S^{\perp})=(u(S))^{\perp}, u^{*}(T^{\perp})=(u^{-1}(T))^{\perp}$$ My question is how to prove $u^{*}(T^{\perp})=(u^{-1}(T))^{\perp}$ and the image of a closed subspace of $W^*$ is a closed subspace of $V^*$.
I've figured out the problem. I think it's not that trivial, so I post the answer. First, note that $V\subset V^{**}$ is a dense subset. Denote by $ X^{\perp(V^{**})}$ the orthogonal of $X$ in $V^{**}$. Then $V\cap X^{\perp(V^{**})}=X^{\perp}$. Denote by $(u^*(T^{\perp}))^{\perp(V^{**})}=R$, $(u^{-1}(T))^{\perp\perp(V^{**})}=S$ . By exercise 1.2.18(ii) (which is easy to prove), we have $V\cap R=V\cap S$ since $T^{\perp\perp}=T$ and $V\cap S=u^{-1}(T)$ by corollary. 1.2.8. If we prove that $R=S$,then we are done since $R^{\perp}=u^*(T^\perp)$ and $S^{\perp}= (u^{-1}(T))^{\perp}$. Note that both $R$ and $S$ are closed in $V^{**}$ and $V$ is a dense subset. Then the closure of $V\cap R$ in $R$ is $R$. But $V\cap R=V\cap S\subset S$ is closed in $ R$. Thus we have $R=S$.