Let $T:V \longrightarrow W$ and consider $\ker(T)^* \simeq V^*/T^*(W^*)$. I am supposed to prove this relationship but I'm having trouble understanding certain components of it.
$\ker(T)^*$ is the dual space of the kernel and $V^*$ is the dual space of $V$. I know that $T^*$ is a dual transformation defined by $T^*:W^* \rightarrow V^*$. I'm not sure what $T^*(W^*)$ is supposed to mean acting on a whole vector space.
And also what does $V^*/T^*(W^*)$ mean? Specifically the divide symbol between $V^*$ and $T^*(W^*)$, is it a coset? If it is can someone please provide the set-builder notation for the elements in that coset? Sorry for asking multiple questions here. While I do understand the components individually I'm struggling to understand the whole. Thanks in advance.
Since $T^*$ is a linear map defined on $W^*$, which is a vector space, then $T^*(W^*)$ is the image of $T^*$. This is a subspace of $V^*$ and so it makes sense to write the quotient $V^*/T^*(W^*)$.
We want to prove that $\ker(T)^* \simeq V^*/T^*(W^*)$. Okay, define $\Phi:V^*\to \ker(T)^*$ by $$\Phi(f)= f\big|_{\ker T}. $$ This is clearly linear.
This is also surjective: given $\widetilde{f}\in \ker(T)^*$, take any extension $f\in V^*$ of $\widetilde{f}$ (which can be obtained, say, by taking a basis $(v_i)_I$ of the kernel, then considering a basis of $V$ of the form $(v_i)_J$ of $V$ with $I\subseteq J$, and defining $f(v_i)=\widetilde{f}(v_i)$ for $i\in I$ and $0$ for $i\in J\setminus I$). Then $\Phi(f)=\widetilde{f}$.
Now, let's compute $\ker\Phi$. We have that $\Phi(f) =0$ if and only if $f\big|_{\ker T}=0$. Ok, if $f\in T^*(W^*)$, then $f= T^*(g)$ for some $g\in W^*$, which says $$f\big|_{\ker T}=g\circ T|_{\ker T} = g\circ 0 =0, $$ so $T^*(W^*)\subseteq \ker \Phi$. On the other hand, if $f\in \ker \Phi$, we must find $g\in W^*$ such that $f = g\circ T$. One can do this as follows: take a basis $(w_k)_K$ of $T(V)$ and consider a basis of $W$ of the form $(w_k)_L$ with $K\subseteq L$. For $k\in K$, we have $w_k = T(x_k)$ for some $x_k\in V$, fix them! Put $g(w_k) = f(x_k)$ for $k\in K$ and $0$ for $k\in L\setminus K$. Done.
Then $T^*(W^*)= \ker \Phi$ and the first isomorphism theorem yields $\ker(T)^* \simeq V^*/T^*(W^*)$, as wanted.