Product of two epimorphisms is an epimorphism I have read that the product of two epimorphisms is always an epimorphism. I am guessing what this means is as follows: Let $C$ be an arbitrary category and $ f : a \to b $ and $ g : c \to d$ epimorphisms in $C$. Then $f \times g : a \times c \to b \times d $ is an epimorphism. I can't seem to prove that $f \times g$ is right cancellative without any assumptions on $C$. Thank you very much for your consideration.
 A: Let me give an example of a coproduct of monomorphisms which is not a monomorphism. Then in the opposite category, you get a product of epimorphisms which is not an epimorphism.
Consider the following category $C$, which I'll call the category of "maybe pointed sets". An object of $C$ has the form $X$, where $X$ is a set, or $(X,a)$, where $X$ is a set and $a\in X$. We call objects of the form $(X,a)$ pointed sets. The morphisms of $C$ are as follows:

*

*An arrow $X\to Y$ is a function $f\colon X\to Y$.

*An arrow $X\to (Y,b)$ is a function $f\colon X\to Y$.

*An arrow $(X,a)\to (Y,b)$ is a function $f\colon X\to Y$ such that $f(a) = b$.

*There are no arrows $(X,a)\to Y$.

Composition and identities are as usual for functions between sets.
The category of maybe pointed sets has coproducts, as follows:

*

*The coproduct of two sets $X$ and $Y$ is their disjoint union (with the obvious inclusions).

*The coproduct of a set $X$ and a pointed set $(Y,a)$ is their disjoint union, pointed by $a$ (and the obvious inclusions).

*The coproduct of two pointed sets $(X,a)$ and $(Y,b)$ is their disjoint union, but with $a$ and $b$ identified, pointed by the identified element $a = b$ (and the obvious inclusions).

Now let $A = \{a\}$, $A' = \{a'\}$, and $B = (\{b\},b)$. We have $A\sqcup A' = \{a,a'\}$ and $B\sqcup B = (\{b\},b)$.
The unique arrows $f\colon A\to B$ and $f'\colon A'\to B$ are monomorphisms (in fact, since $A$ is a singleton, any two arrows with the same domain and with codomain $A$ are already equal, and the same for $A'$). But their coproduct $(f\sqcup f')\colon A\sqcup A'\to B\sqcup B'$, defined by $f(a) = f(a') = b$, is not a monomorphism.
Indeed, let $C = \{c\}$, and consider the arrows $g,g'\colon C\to A\sqcup A'$ defined by $g(c) = a$ and $g'(c) = a'$. We have $g\neq g'$, but $(f\sqcup f')\circ g = (f\sqcup f')\circ g'$. So $(f\sqcup f')$ is not a monomorphism.
