What I want to show: Let $f:V\to V$ be an endomorphism of the finite-dimensional $K$-vector space $V$. Let $U\subseteq V$ be a $K$-vector subspace that's invariant under the linear map $f$ (so $U$ satisfies: $f(U)\subseteq U$). Furthermore let $$\overline{f}: V\backslash U\to V\backslash U, \quad v+U\mapsto f(v)+U$$ be the induced endomorphism of the quotient space $V\backslash U$ and $f|_U: U\to U$ the restriction of $f$ to $U$. Then the following equation is satisfied (this is what I want to show): $$\det(f)=\det(\overline{f})\det(f|_U).$$
My thoughts: If I can show that $f(v)=f|_U(\overline{f}(v))$ holds for all $v\in V$, then $f$ would be equal to $f|_U\circ\overline{f}$. Then the matrix that corresponds to the linear map $f$ (let's call it $A_f$) would be equal to $A_{f|_U\circ\overline{f}} = A_{f|_U}\cdot A_\overline{f}$, where $A_{f|_U}$ is the matrix corresponding to $f|_U$ and $A_\overline{f}$ is the matrix corresponding to $\overline{f}$. So after that I could just use the identity for the determinant of the product of two matrices and get $$\det(A_f)=\det(A_{f|_U\circ\overline{f}})=\det(A_{f|_U}\cdot A_\overline{f})=\det(A_{f|_U})\det(A_\overline{f})$$ which is exactly what I wanted to show because the determinant of a linear map is equivalent to the determinant of its corresponding matrix. So in the end I would get $\det(f)=\det(f|_U)\det(\overline{f})=\det(\overline{f})\det(f|_U)$ and the proof would be finished.
Where I need help: I don't know how I could show that $f(v)=f|_U(\overline{f}(v))$ holds for all $v\in V$. Maybe showing $f(u)=\overline{f}(f|_U(u))$ works too? But since $f|_U$ is the restriction of $f$ to $U$ I would just proof that it holds for all $u\in U$ and that wouldn't be enough, I think.
I would very much appreciate if someone could help me out with this!