How to describe the vector space $\left( \frac{Z}{Im(T)} \right)^*$?

Question

Just as a preface: I am not looking for an answer I just want help describing what the vector space looks like so that I can work on solving the problem.

We have a linear transformation from $T: V \to Z$ where $V$ and $Z$ are potentially infinite-dimensional vector spaces.

I have to describe an isomorphism from $\text{Ker}(T^*) \to \left( \frac{Z}{Im(T)} \right)^*$ where the asterisk denotes the dual space.

I have a good understanding of $\text{Ker}(T^*)$. $$T^* : Z^* \to V^*$$ where $T^*$ sends functionals $f \in Z^*$ to the zero functional $f_0 \in V^*$.

But, I have no idea what $\left( \frac{Z}{Im(T)} \right)^*$ looks like. The image of $T$ is exactly $Z$ so does this mean $\left( \frac{Z}{Im(T)} \right)^* = \left( \frac{Z}{Z} \right)^*$?

Answer 1

Using the quotient map, define an isomorphism from $\left(\frac{Z}{Im(T)}\right)^*$ to $\{f \in Z^* : f(Im(T))=0\}$.

Then it is immediate that it is isomorphic to $Ker(T^*)$ by the definition of $T^*$.

Answer 2

You write that $T$ is a linear transformation $V\to Z$, and apparently that is all you know about it.

You can't then conclude that "the image of $T$ is exactly $Z$", since that would mean that $T$ is bijective, and you have no guarantee that this is the case.

Remember that by definition the image of $T$ is $\{Tv\mid v\in V\}$ -- when writing $T:V\to Z$ you're just told that $Tv\in Z$ always, not that all members of $Z$ have the form $Tv$ for some $v$.

$Z$ is the codomain of $T$.

(Some authors say "range" instead of "image", and may then say "image" for "codomain" -- but don't let that usage confuse you).

---

Thus, $Z/\operatorname{Im}T$ is a vector space whose elements are equivalence classes of the form $ \overline z = \{ z+Tv \mid v\in V \}$. If you have a linear functional $f$ on this quotient space, you can derive a linear functional $g$ on $Z$ by $$ g(z) = f(\overline z) $$ which will then have the property that $g(Tv)=0$ for all $v$. On the other hand, every such $g$ will give rise to an $f$ in this way ...