When is $p_1 \times p_2: V\rightarrow V/U_1 \oplus V/U_2, v\longmapsto (p_1(v),p_2(v))$ surjective? Let $K$ be a field, $U_1, U_2 \subset V$ vector subspaces of a $K$-vector space. Considering the following canonical $K$-linear transformations:
$$p_1: V \rightarrow V/U_1$$ $$v \mapsto v+U_1$$ $$ $$
$$p_2: V \rightarrow V/U_2$$ $$v \mapsto v+U_2$$
Consider 
$$p_1 \times p_2: V\rightarrow V/U_1 \oplus V/U_2$$ $$v\longmapsto (p_1(v),p_2(v))$$
$$ $$
Show that $p_1 \times p_2$ is an epimorphism exactly when $U_1+U_2=V$.
Consider $V=\mathbb{R^3}$. Find $U_1 \neq U_2$ such that $p_1 \times p_2$ is neither injective nor surjective.
I know that all $v \in V/U_1 \oplus V/U_2$ are of the form $(v+U_1, v+U_2)$, but I do not understand why $U_1, U_2$ must be complementary for $p_1 \times p_2$ to be surjective. 
Also, the map is not surjective if $U_1+U_2\neq V$, but it is not injective when for $v, w \in V$ with $v \neq w$ $(p_1 \times p_2)(v)=(p_1 \times p_2)(w)$. Thus it implies that $p_1(v)=p_1(w)$ and $p_2(v)=p_2(w)$. In other words $v+U_1=w+U_1$ and $v+U_2=w+U_2$ and I know that it implies that $v-w \in U_1$ and $v-w \in U_2$, thus $v-w \in U_1 \cap U_2$ for all $v,w \in V$, but I do not know what that means for $U_1$ and $U_2$.
 A: Elements of $V/U_1 \oplus V/U_2$ are of the form $(v_1+U_1, v_2+U_2)$ for some $v_1,v_2 \in V$. Note that you made an error by using the same $v$ on both sides, which would be the image of $p_1 \times p_2$ which a priori is contained in $V/U_1 \oplus V/U_2$. But this problem concerns the setting where the image is all of $V/U_1 \oplus V/U_2$ (surjectivity).
To summarize, the image of $p_1 \times p_2$ is $$\{(v+U_1,v+U_2):v \in V\}$$
while the codomain is
$$V/U_1 \oplus V/U_2 := \{(v_1+U_1,v_2+U_2):v_1,v_2 \in V\}.$$

Suppose $U_1+U_2=V$. Then for any $(v_1+U_1, v_2 + U_2) \in V/U_1 \oplus V/U_2$, we claim there exists $v$ such that $(p_1 \times p_2)(v) := (v+U_1,v+U_2) = (v_1+U_1,v_2+U_2)$ so $p_1 \times p_2$ is surjective. To prove this claim, write $v_1-v_2 = u_1+u_2$ where $u_1 \in U_1$ and $u_2 \in U_2$ using the condition $U_1+U_2=V$. Then $v_1-u_1=v_2+u_2$; let this quantity be $v$.
For the converse, suppose $p_1 \times p_2$ is surjective. For any $v$, we can consider $(v+U_1,0+U_2) \in V/U_1 \oplus V/U_2$. Surjectivity implies there exists some $v'$ such that $(p_1 \times p_2)(v')=(v'+U_1,v'+U_2)=(v+U_1, 0+U_2)$. Equating each component gives $v-v' \in U_1$ and $v' \in U_2$, so $v=(v-v')+v'$ shows $v \in U_1+U_2$.

For the other problem, try $\{0\} \subsetneq U_1 \subsetneq U_2 \subsetneq \mathbb{R}^3$. Then $U_1+U_2=U_2 \ne \mathbb{R}^3$ so it is not surjective. Also, if you choose $v,w \in U_1$, then both map to $(0+U_1,0+U_2)$ so it is not injective.
