Proof of $\dim({\rm Ker}(g\circ f)) \leq \dim({\rm Ker}(f)) + \dim({\rm Ker}(g))$

Question

This is an exercise that appeared in an Algebra II final exam from some years ago I found while studying a few days ago:

Let $U, V, W$ be finite-dimensional K-Vector spaces and $f:U \rightarrow V$, $g:V \rightarrow W$ linear transformations. Prove that:

$$\dim({\rm Ker}(g\circ f)) \le \dim({\rm Ker}(f)) + \dim({\rm Ker}(g)) $$

HINT: Consider the restriction of $f$ to the space ${\rm Ker}(g\circ f)$

So far I only got that

$$\dim({\rm Ker}(g\circ f))=\dim({\rm Im}(f\restriction_{{\rm Ker}(g\circ f)}))$$

using the hint given, but I've been thinking for a while and I don't know how to link that to ${\rm Ker}(f)$ and ${\rm Ker}(g)$. I thought it might be related to the dimension of the direct sum.

I also found an answered question for this same exact proof but it has a different approach, so I was wondering if someone could give me some guidance for the proof using the hint given.

Thanks in advance.

Answer 1

If $f’ \colon \ker(g \circ f) \to V$ is the restriction of $f$ to $\ker(g \circ f)$, then $$ \ker(f’) = \ker(g \circ f) \cap \ker(f) = \ker(f) \\ \& \quad \text{im}(f’) = f(\ker(g \circ f)) \subseteq \ker(g). $$ What does the rank-nullity theorem say when applied to $f'$?