Let $\mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A\_i)\_{i\in I}$ and $(B\_i)\_{i\in I}$ in $\mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A\_i)\_{i\in I}$ and of $(B\_i)\_{i\in I}$, respectively. If $f\_i:A\_i\to B\_i$ are morphisms for all $i\in I$, then there exists a unique morphism $f:A\to B$, which is the coproduct of $(f\_i)\_{i\in I}$.


I know the following results. The morphism $f$ need not be monic, even if all $f\_i$ are monic. If $\mathbf{C}$ is an abelian category and if all $f\_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.



> 
> **Question I.** In general (where $\mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f\_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
> 
> 
> **Question II.** If $\mathbf{C}$ is an abelian category and if all $f\_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
> a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
> 
> 
> 



Is a categorical coproduct of epimorphisms (monomorphisms) always an epimorphism (a monomorphism)?
Question 1: Yes. The $I$-coproduct-functor $\bigsqcup\_I\colon\prod\_{i\in I}\mathbf{C}\to\mathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $\Delta\_{\mathbf{C}}^I\colon \mathbf{C}\to\prod\_{i\in I}\mathbf{C}$), hence always preserves epimorphisms.


Question 2: No, in general (even if $\mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called *satisfying axiom AB4*. See [nlab article on Grothendieck axioms](https://ncatlab.org/nlab/show/additive+and+abelian+categories). Some counterexamples were discussed on SE; see, for example [Zhen Lin's answer on MSE](https://math.stackexchange.com/questions/674482/example-of-epimorphisms-such-that-the-product-is-not-an-epimorphism-in-the-categ). The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).


Regarding the changes in your question after edits:


In the situation when $\mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact: 
$$
(g\circ f=h\circ f)\Rightarrow(g\circ f\circ s\_i=h\circ f\circ s\_i)\Rightarrow(g\circ s\_i\circ f\_i=h\circ s\_i\circ f\_i)\Rightarrow(g\circ s\_i=h\circ s\_i)\Rightarrow(g\circ h),
$$
where $g$ and $h$ are arbitrary morphisms of $\mathbf{C}$ with domain $B$ and $s\_i$, $i\in I$, are canonical injections of coproducts.


As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.


