Is taking cokernels coproduct-preserving? Let $\mathcal{A}$ be an abelian category, $A\,A',B$ three objects of $\mathcal{A}$ and $s: A\to B$, $t: A' \to B$ morphisms.
Is the cokernel of $(s\amalg t): A\coprod A'\to B$ the coproduct of the cokernels of $s$ and $t$?
In case it's wrong: Is it true if we restrict $s$ and $t$ to be monomorphisms?
 A: Colimits preserve colimits, so colimits do preserve coproducts, and cokernels are colimits. However, this means something different than what you suggest.
Usually, $s \coprod t$ is used to mean the morphism $A \coprod A' \to B \coprod B$; with this meaning, we do have
$$ \text{coker}(s \coprod t) = \text{coker}(s) \coprod \text{coker}(t)$$
If I let $(s,t)$ denote the morphism $A \coprod A' \to B$, then if I've not made an error, what we do have is a pushout diagram
$$ \begin{matrix}
B &\to& \text{coker}(s)
\\\downarrow & & \downarrow
\\\text{coker}(t) &\to& \text{coker}(s,t)
\end{matrix} $$
or equivalently, we have an exact sequence
$$ B \to \text{coker}(s) \oplus \text{coker}(t) \to \text{coker}(s,t) \to 0 $$
A: I don't think it works, not even if $s,t$ are both mono.
For a concrete counterexample, take $\mathcal{A}$ to be the category of $K$-vector spaces, $A=A'=K$ and $B=K^2$. If $s,t$ are both non-zero (hence injective), their cokernels are both one-dimensional, so that the coproduct $\text{coker}(s) \coprod \text{coker}(t) \cong K^2$ has dimension two.
On the other hand, since both $s$ and $t$ are non-zero, the induced map $s \coprod t: K^2 \rightarrow K^2$ is not the zero map, hence its cokernel has dimension $< 2$.
Edit: Actually, for morphisms $s: A \rightarrow B$, $t: A' \rightarrow B'$ the morphism $s \coprod t$ usually denotes the canonical map 
\begin{split}
A \coprod A' \rightarrow B \coprod B'
\end{split}
as noted by Hurkyl. If $B=B'$, then the morphism $s \coprod t$ is something else than the morphism $A \coprod A' \rightarrow B$ given by the universal property of the coproduct, and it seems you were asking about the latter, while using notation usually reserved for the former.
