Well-defined, surjective map between two quotient spaces (Linear Algebra - beginner level)

Question

once again I am working on a proof in Linear Algebra. I have tried to write it in a formally correct way. At a certain point I don't know how to proceed, although I have an underlying idea for how to do it. I marked this with a line.

Statement:

Let $U \subset V \subset W$ be vector spaces and denote the elements of the quotient spaces $ W/U $ and $ W/V $ by $[w]_U$ and $[w]_V$ for $w \in W$.

Show that $\Psi : W/U \rightarrow W/V$ , $[w]_U \mapsto [w]_V $ is well defined and surjective.

Proof:

Begin by showing that the relation $\Psi$ is well defined, which would allow us to classify it as a function. First, show that every element $[w]_U \in W/U$ is in relation to some $[w]_V \in W/V$. Secondly, show that $\Psi$ is independent of the choice of representative. This is important, since if $\Psi$ is to be a function, there can be no element in its domain which gets mapped to more than one $[w]_V \in W/V$.

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Idea for continuing this proof:

I want to define $\Psi$ by cases depending on $w \in W$. So for example, if $w \in U$, every $[w]_U = [0]_U$ could be mapped to $[0]_V$ (By choosing the representatives in a fitting way, it would be possible to say that $[0]_U =[0]_V$). The other cases could be $w \in$ V\U and $w \in$ W\V. In order to do this I first have to somehow fix the representatives $[w]_U$ and $[w]_V$, so that the second property of a well defined function in the above beginning of the proof can be fulfilled. At this point I'm not sure how to continue.

Answer 1

$\Psi$ is well-defined:

Let $[w]_U, [w']_U \in W/U$ such that $[w]_U= [w']_U$. We have

$$[w]_U = [w']_U \implies w - w' \in U \subseteq V \implies w- w' \in V \implies [w]_V = [w']_V$$

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$\Psi$ is surjective:

Let $[w]_V \in W/V$. Then $[w]_U \in W/U$ and we have

$$\Psi([w]_U) = [w]_V$$

by definition of $\Psi$.

Answer 2

If for $w_1,w_2\in W$ You have $[w_1]_U=[w_2]_U$ then $w_1-w_2\in U$ and because $U\subset V$ this implies $w_1-w_2\in V $ and thus $[w_1]_V=[w_2]_V$,i.e. $\Psi$ is well defined. For $[w]_V$ You have $\Psi([w]_U)=[w]_V$ and thus $\Psi$ is surjective.