So, here's the result that I'm trying to prove:
The map given by:
$$\pi: V \to V/U$$ $$v \to v+U$$
is an epimorphism with $Ker(\pi) = U$.
Proof Attempt:
We need to show that this is a linear map that is surjective. We begin by proving linearity first.
Let $v_1,v_2 \in V$. Then:
$\pi(v_1+v_2) = (v_1+v_2) + U = (v_1 + U) + (v_2 + U) = \pi(v_1) + \pi(v_2)$
That proves the first condition for linearity. Let $\lambda \in F$ and $v \in V$. Then:
$\pi(\lambda v) = \lambda v + U = \lambda \cdot (v + U) = \lambda \cdot \pi(v)$.
That proves the second condition for linearity. This proves that the given function is a linear map.
Then, we have to prove surjectivity. Clearly, $\pi(V) \subset V/U$. Let $v+U \in V/U$ for some $v \in V$. However, that just means that $\pi(v) = v+U$, so $v+U \in \pi(v)$. This shows that $V/U \subset \pi(V)$ and, thus, proves that $\pi(V) = V/U$.
Finally, we have to show that $Ker(\pi) = U$. We do this by mutual containment. Let $v \in Ker(\pi)$. Then, $\pi(v) = v + U = U$ and that is possible only when $v \in U$. That shows that $Ker(\pi) \subset U$.
Then, let $v \in U$. Clearly, $v+U = U$. However, that means that $\pi(v) = v+U = U$ and that implies that $v \in Ker(\pi)$. That shows that $U \subset Ker(\pi)$. Thus, $Ker(\pi) = U$.
Can someone have a look at my proof above and see if it makes sense or not? I'm kind of wonky on quotient vector spaces because my book mentions them only very briefly and doesn't give much of an intuition for them beyond the basic definitions. Also, would there be a way to shorten the proof above? Thanks in advance!