Prove that $ \dim(\ker(g \circ f)) = \dim(\ker(f)) + \dim(\ker(g) \cap \operatorname{im}(f)). $ What is the error in this approach?

Question

Let $f : U \to V$ and $g : V \to W$ be linear transformations on the vector spaces $U$, $V$, and $W$ respectively. Then Prove that :

$$ \dim(\ker(g \circ f)) = \dim(\ker(f)) + \dim(\ker(g) \cap \operatorname{im}(f)). $$

Attempt: It's clear that $$\ker(g \circ f))= \ker f +\{\operatorname{im} f \cap \ker(g)\}$$

Then $$\dim \ker(g \circ f))= \dim \big (\ker f +\{\operatorname{im} f \cap \ker(g)\} \big ) \tag{1} $$

If $\ker f \bigcap \operatorname{im}f=\emptyset$, then , $(1)$ becomes an external direct product and the dimension adds up. Otherwise, $\dim \ker(g \circ f)) \le \dim \big (\ker f)+\dim \big(\{\operatorname{im} f \cap\ker(g)\} \big )$

Could someone please pin point the error. Thanks a lot for the help.

Answer 1

The thing that is clear is that $\textrm{ker}(g\circ f)=f^{-1}(\textrm{ker}(g))$.

Now, $f$ induces an isomorphism $\varphi:U/\textrm{ker}f\to \textrm{Im}f,$ and since $\textrm{ker}f\subseteq f^{-1}(\textrm{ker}g),$ we have \begin{align} dim(f^{-1}(\textrm{ker}(g))&=dim(f^{-1}(\textrm{ker}(g)\cap \textrm{Im} f))\\ &=dim(\textrm{ker} f)+dim(f^{-1}(\textrm{ker}(g)\cap \textrm{Im} f)/\textrm{ker}f)\\ &= dim(\textrm{ker} f)+dim(\varphi^{-1}(\textrm{ker}(g)\cap \textrm{Im}f))\\ &=dim(\textrm{ker} f)+dim(\textrm{ker}(g)\cap \textrm{Im}f), \end{align} since $\varphi$ is an isomorphism.