A question on kernel and image of a composition of linear transformation.

Question

Let $V,W,U$ be vector spaces over $\mathbb{F}$ such that $V$ and $W$ are finite dimensional. Let $f:V\to W$ and $g:W\to U$ be linear transformations over $\mathbb{F}$.

a) Let $h:f(V)\to U$ such that $h(w):=g(w)$ for all $w\in f(V)$. How are Im$(g\circ f)$ and Im$(h)$ related and why? How are ker$(h)$ and ker$(g)$ related and why?

b) Show that rank$(f)$$\leq$rank$(g\circ f)$+nul$(g)$.

For item a, what I did was to use the fact that images are subspaces of the codomain of a linear transformation hence it kinda sounds like the images are the same, but I actually have no idea how to move on nor if the why I think of it is correct. For item b, I haven't had the luck yet. Can I ask for help for both items? A hint would be so appreciated, I still want to work out the proof on my own. I just want a headstart. Thank you.

Answer 1

For a, your guess that the images are the same is correct. Keeping in mind the definition of the image of a linear transformation, it suffices to note that for every $w \in f(V)$, $h(w)$ can be written in the form $g(f(v))$ for some vector $v \in V$.

In order to relate the kernel of these two maps, note that if $h(w) = 0$ holds for some $w \in f(V) \subset W$, then it follows that $g(w) = 0$.

For b, recall that the rank of $f$ is the dimension of $f(V)$. Now, apply the rank-nullity theorem to the map $h$.