Let $A_0 + A_1$ be the coproduct of $A_0$ and $A_1$. Suppose the canonical injections $i_0:A_0 \rightarrow A_0 + A_1$ and $i_1:A_1 \rightarrow A_0 + A_1$ are both monomorphisms. A point of an object is an arrow into it from the terminal object, $1$. If $x:1 \rightarrow A_0 + A_1$ is a point of the coproduct then is it true that either $x$ is a member of $i_0$ (in the sense that there exists $k_0$ such that $x = i_0 \circ k_0$) or that $x$ is a member of $i_1$ (in the sense that there exists $k_1$ such that $x = i_1 \circ k_1$) ? If so, what is the proof ?
In other words, I am wondering whether each point of the coproduct has to correspond to a point of one of the objects involved in the coproduct. If it helps, one may assume the category is a topos.
No. Probably the simplest counterexample is the topos $\mathbf{Set}^2$, whose objects are ordered pairs of sets and whose morphisms are ordered pairs of functions. In this topos, the terminal object is a pair of singletons, which I'll wrwte as $(\{a\},\{b\})$. To help keep the notation simple, I'll take a second copy of this terminal object, say $(\{c\},\{d\})$, so I can write the coproduct of two copies of the terminal object as $(\{a,c\},\{b,d\})$ with the obvious injections. One of the points of this coproduct is the morphism $(f,g)$ from $(\{a\},\{b\})$ to $(\{a,c\},\{b,d\})$ where $f:\{a\}\to\{a,c\}$ sends $a$ to $a$ and $g:\{b\}\to\{b,d\}$ sends $b$ to $d$. (So the first component "comes from" the first injection and the second component "comes from" the second injection.) This point $(f,g)$ does not factor through either of the coproduct injections.