If $F(U)⊂ U$, then $ϕ : V/U → V/U , v + U → f(v) + U $, but why?

Question

There is given vector space $V$ through a field $K$ and f: $V → V$ is an Endomorphism. Furthermore, $ U ⊂ V $ is a vector subspace and the following is true : $f(U)⊂ U$

Furthermore there is a mapping $ϕ : V/U → V/U $

In the University our professor sad that if $f(U)⊂ U$ is true, it automatically means $\phi([v]) = f(v) +U$. But I do not see how, I do not undestrand why $\phi([v]) = f(v) +U$ follows $f(U)⊂ U$.

Can somebody please help me out?

Answer 1

You seem to have things mixed up. Here's what I think your professor said:

We're given a vector space $V$ over $K$, an endomorphism $f: V \to V$, and an invariant subspace $U$ (i.e. a subspace $U$ such that $f(U) \subset U$).

Given this information, we can define a mapping $\phi:V/U \to V/U$ by $$ \phi(v + U) = f(v) + U $$

Notably, the $\phi$ defined here is often called the map which is induced by $\phi$ on $V/U$. If $f(U) \not\subset U$, then we no longer have the required setup, which is to say that there is no map $\phi$.

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So, assuming the above makes sense to you, you might ask "why was it important to have $f(U) \subset U$?" Of course, we can always define a map $\psi: V \to V/U$ by $$ \psi(v) = f(v) + U $$ However, in order to make this into a map over the quotient $V/U$, we would need to make sure that the map is well defined. In particular, it is necessary to have $$ v + U = w + U \implies f(v) + U = f(w) + U $$ otherwise, the definition $\phi(v + U) = f(v) + U$ becomes ambiguous. By linearity, we can rearrange the above equation: $$ (v - w) + U = 0 + U \implies f(v - w) + U = U \quad \forall v,w \in V $$ or in other words, $$ v-w \in U \implies f(v-w) \in U \quad \forall v,w \in V $$ but this is equivalent to the statement that $f(U) \subset U$.