Showing that the transformation from dual of a quotient space to the kernel of the dual transformation is injective

Question

Let $T: V \rightarrow W$, I am trying the prove the following transformation is injective: $G:(W/T(V))^* \rightarrow \ker(T^*)$. $G$ is defined as $$G(h(w)) = h(w + T(V))$$ where (I'm assuming) $h \in (W/T(V))^*$ and $w \in W$. To prove it is injective I have to show that if $G(h(w)) = G(f(w))$ then $h(w) = f(w) = 0$ as in the zero function of of $(W/T(V))^*$. I have two questions, first I'm kind of confused as to the transformation's definition. Because it seems that $h(w+T(V))$ is an element of $(W/T(V))^*$ rather than $\ker(T^*)$, am I missing something here? Also, how would I approach showing injectivity? Thanks in advance, and please let me know if something is unlcear.

Answer 1

We have $G:(W/T(V))^* \rightarrow \ker(T^*)$. You say that $G$ is defined as $G(h(w)) = h(w + T(V))$, but $h$ is a map $h:W/T(V)\to \Bbb K$, so we must be careful with notation. Better to write $G(h)(w)$ instead of $G(h(w)) $ (that last expression does not compile!). We'll have that $G(h)\in W^*$, which in turn is defined as $G(h)(w) = h(w+T(V))$, ok. We really have that $G(h)\in \ker(T^*)$ because $T^*(G(h)) = G(h)\circ T$ is the zero map, since $$(G(h)\circ T)(v) = G(h)(T(v)) = h(T(v)+T(V))=h(0+T(V))=0,$$so the codomain is correct, fine.

Now, to see that $G$ is injective. Take $h\in \ker G$. So $G(h)$ is the zero map, and for all $w\in W$ we have $h(w+T(V)) = 0$. This implies that $h=0$ because every element in $W/T(V)$ is of the form $w+T(V)$ for some $w\in W$.