$\pi(v)=x+y$ implies $v=v_1+v_2,\; \pi(v_1)=x,\pi(v_2)=y$

Question

Let $V$ be a vector space over some field $k$, let $U\subset V$ be a subspace of $V$ and consider the quotient space $V/U$, along with the projection $\pi:V\to V/U,\, x\mapsto x+U$.

Does it hold that $\pi(v)=x+y$ implies $v=v_1+v_2$ with $ \pi(v_1)=x,\pi(v_2)=y$?

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My approach: $x+y=\pi(\tilde{x})+\pi(\tilde{y})=\pi(\tilde{x}+\tilde{y})$ for $\tilde{x},\tilde{y}\in V$. How do I continue from here in order to show that $v$ splits in the above way?

Answer 1

The answer is positive.

Suppose that $\pi(v) = x+y$. As $x \in V/U$ and $\pi$ is surjective, it exists $v_1^\prime \in V$ such that $\pi(v_1^\prime)=x$. Similarly for $y$ with $v_2^\prime$.

We then have $\pi(w)=0$ where $ w = v-(v_1^\prime +v_2^\prime)$. Which means that $ w $ belongs to $U$.

So $v=(v_1^\prime+w)+v_2^\prime$ and $v_1=v_1^\prime+w$, $v_2=v_2^\prime$ solve the problem as $\pi(v_1^\prime) =\pi(v_1)$.