If $f: U \rightarrow V$ and $g: V \rightarrow W$ are linear transformations between vector spaces over a field $K$ such that $ g \circ f$ is an isomorphism, then $V = \operatorname{Im}f \oplus \operatorname{Ker} g$.
I tried this way: since $g\circ f\colon U\to W$ is an isomorphism, I can set $h\colon W\to U$ to be its inverse; then I have $V$ back to $V$ with $\varphi=f\circ h\circ g$. If $v\in V$, so $v=\varphi(v)+(v-\varphi(v))$, then $V = \operatorname{Im} f \oplus \operatorname{Ker} g$.
Is it correct? Does there exist a direct or more simple proof of this?
Some help please, thanks for your time.
The idea is correct. Indeed $\varphi(v)=f(h(g(v)))\in\operatorname{Im}f$. Moreover, $$ g(v-\varphi(v))=g(v)-(g\circ f\circ h\circ g)(v)=g(v)-g(v)=0 $$ since $g\circ f\circ h$ is the identity.
This proves that $V=\operatorname{Im}f+\operatorname{Ker}g$, but you also need to show that $$ \operatorname{Im}f\cap\operatorname{Ker}g=\{0\} $$ You should be able to do it.