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How do you see that if $F\colon X\to Y$ and $G\colon Y\to Z$ are two linear maps between the vector spaces $X$ and $Y$, then $$\dim(\ker(G\circ F)) \leq \dim(\ker(F)) + \dim(\ker(G))$$?

I see that $\ker(G\circ F) \equiv F^{-1} (G^{-1}( \{0\} ))$ but I'm not exactly sure how this helps. I can also see that $\ker(G\circ F) \supseteq \ker(F)$ in $X$, that is, there is a larger chunk of $X$ (larger than $\ker(F)$) which can get mapped into $\ker(G)\subseteq Y$.

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As you wrote, $\ker(G \circ F) = F^{-1}(\ker G)$. You can prove more generally that if $U \leq Y$ is a subspace then

$$ \dim F^{-1}(U) \leq \dim U + \dim \ker F. $$

To see this, consider $F|_{F^{-1}(U)} \colon F^{-1}(U) \rightarrow U$ and apply the rank-nullity theorem.

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Here is a method that applies when we are considering bounded linear maps between (possibly infinite-dimensional) Banach spaces. In this context, since the space $X$ may be infinite-dimensional, we cannot apply the rank-nullity theorem.

If $\dim\ker(F) = \infty$, then there is nothing to show, so suppose that $\dim\ker(F) < \infty$. Since $\ker(F) \subset \ker (G\circ F)$, and $\ker(G\circ F)$ is a closed subspace of the Banach space $X$ (hence a Banach space in its own right), there is a closed complement $C$ for $\ker(F)$ in $\ker(G\circ F)$, i.e., a closed subspace $C$ of $\ker(G\circ F)$ such that $$ \ker(G\circ F) = \ker (F)\oplus C. $$ Now, $F|_C\colon C\to \ker(G)$ is a vector space isomorphism onto its image $F(C)$, so $\dim (C) = \dim F(C) \le \dim \ker (G)$ (where the inequality holds as an inequality of extended real numbers). Hence we have $$ \dim\ker (G\circ F) = \dim \ker(F) +\dim(C)\le \dim \ker (F) + \dim \ker (G), $$ as desired.

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