Let $V,W, U$ be vector spaces over $F$ where $V$ and $W$ are finite dimensional. Let $f: V \to W$ and $g: W \to U$ be linear transformations over $F$. We need to prove that:
$$ \textbf{rank} (f) \leq \textbf{rank}(g \circ f) + \textbf{nullity} (g)$$.
Can you guys please check if my proof is valid?
This is what I have:
What I did is to first restrict $g: W \to U$ to $g': f(V) \to U$. Thus we can say by the Rank-Nullity Theorem that $$ \textbf{dim}f(V) = \textbf{rank} (g') + \textbf{nullity} (g').$$ Notice that $\textbf{dim}f(V) = \textbf{rank}(f)$ and that $\textbf{rank}(g')=\textbf{rank}(g \circ f)$ and so we further have $$ \textbf{rank}(f) = \textbf{rank}(g \circ f) + \textbf{nullity}(g').$$
Now, $\textbf{ker}(g') = \{y \in f(V) | g(y) = 0 \}$. Since $f(V)$ is a subspace of $W$, then that means $\textbf{ker}(g')$ is a subspace of $\textbf{ker}(g)$. This implies that $\textbf{nullity} (g') \leq \textbf{nullity} (g)$. Finally, $$ \textbf{rank} (f) \leq \textbf{rank}(g \circ f) + \textbf{nullity} (g)$$.