What is the relation between the cokernel with the kernel of the dual map of a linear transformation? I am studying linear algebra and I am in front on questions like: What is the relation between the kernel of a linear map and the cokernel of the dual map?
What is the relation between theese objects and the existence of solution for the equation $f(x) = y$?
Thanks in advance!
 A: I'll give a partial answer, since I also have no clue of the relation with the equation $f(x) = y$ as it is stated.
Let $T\colon V \to W$ be a linear map and $T^\ast\colon W^\ast \to V^\ast$ be the dual map. I claim that if the spaces have finite dimension, then ${\rm coker}(T) \cong \ker(T^\ast)$. 
First of all, let's check that $\ker(T^\ast) = {\rm Ann}({\rm Im}(T))$. Just note that: $$w^\ast \in \ker(T^\ast) \iff T^\ast(w^\ast) = 0 \iff  w^\ast \circ T = 0 \iff w^\ast\bigg|_{{\rm Im}(T)} = 0 \iff w^\ast \in {\rm Ann}({\rm Im}(T)).$$
In finite dimension, we have $$\dim {\rm Im}(T) + \dim{\rm Ann}({\rm Im}(T)) = \dim W,$$ you can check this in page $124$ of Flavio Coelho's linear algebra book, which I'm sure you know (hue BR!).
Putting all of this together, we have: $$\dim{\rm coker}(T) = \dim \frac{W}{{\rm Im}(T)} = \dim W - \dim {\rm Im}(T) = \dim {\rm Ann}({\rm Im}(T)) = \dim \ker(T^\ast),$$ so we have the isomorphism. I believe strongly that we'll have problems in infinite dimension.
